Mechanics of Fluids , Fourth Edition

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Author: Merle C. Potter | David C. Wiggert | Bassem H. Ramadan

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Conversion Factors to SI Units

English

SI

SI symbol

To convert from English to SI multiply by

To convert from SI to English multiply by

Acceleration foot/second squared

meter/second squared

m/s2

0.3048

3.281

cm2 m2 ha

6.452 0.09290 0.4047

0.1550 10.76 2.471

Area square inch square foot acre

square centimeter square meter hectare Density

slug/cubic foot

kg/m3

kilogram/cubic meter

515.4

1.94 ⫻ 10⫺3

Flow Rate cubic foot/second cubic foot/second

cubic meter/second liter/second

m3/s L/s

0.02832 28.32

35.32 0.03532

N N

4.448 4448

0.2248

cm m km

2.54 0.3048 1.6093

0.3937 3.281 0.6214

kg kg

0.4536 14.59

2.205 0.06854

0.7457 1.356 0.2929

1.341 0.7376 3.414

Force pound kip (1000 lb)

newton newton Length

inch foot mile

centimeter meter kilometer Mass

pound mass slug

kilogram kilogram

Power, Heat Rate horsepower foot-pound/sec Btu/hour

kilowatt watt watt

kW W W

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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Conversion Factors to SI Units (continued)

English

SI

SI symbol

To convert from English to SI multiply by

To convert from SI to English multiply by

Pressure pound/square inch pound/square foot feet of H2O inches of Hg

kilopascal kilopascal kilopascal kilopascal

kPa kPa kPa kPa

6.895 0.04788 2.983 3.374

0.1450 20.89 0.3352 0.2964

Temperature Fahrenheit Fahrenheit

Celsius kelvin

°C K

5/9(°F ⫺ 32) 5/9(°F ⫹ 460)

9/5 ⫻ °C ⫹ 32 9/5 ⫻ K ⫺ 460

1.356 0.1130

0.7376 8.85

0.3048 0.4470 1.609

3.281 2.237 0.6215

47.88 0.09290

0.02089 10.76

16.387 0.02832 0.003785 3.785

0.06102 35.32 264.2 0.2642

1.356 1.054 0.000293 29.3

0.7376 0.9479 3413 0.03413

Torque pound-foot pound-inch

N⭈m N⭈m

newton-meter newton-meter

Velocity foot/second mile/hour mile/hour

meter/second meter/second kilometer/hour

m/s m/s km/h Viscosity, Kinematic Viscosity

pound-sec/square foot square foot/second

newton-sec/square meter square meter/second

N ⭈ s/m2 m2/s Volume

cubic inch cubic foot gallon gallon

cubic centimeter cubic meter cubic meter liter

cm3 m3 m3 L Work, Energy, Heat

foot-pound Btu Btu therm

joule kilojoule kilowatt-hour kilowatt-hour

J kJ kWh kWh


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Mechanics of Fluids


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Properties of Water

Temperature (°C) 0 5 10 15 20 30 40 50 60 70 80 90 100

Density r, (kg/m3)

Specific weight g, (N/m3)

Viscosity m, (N ⭈ s/m2)

999.9 1000.0 999.7 999.1 998.2 995.7 992.2 988.1 983.2 977.8 971.8 965.3 958.4

9809 9810 9807 9801 9792 9768 9733 9693 9645 9592 9533 9470 9402

1.792 ⫻ 10⫺3 1.519 1.308 1.140 1.005 0.801 0.656 0.549 0.469 0.406 0.357 0.317 0.284 ⫻ 10⫺3

Kinematic viscosity n, (m2/s)

Bulk modulus B, (Pa)

Surface tension s, (N/m)

Vapor pressure, (kPa)

1.792 ⫻ 10⫺6 1.519 1.308 1.141 1.007 0.804 0.661 0.556 0.477 0.415 0.367 0.328 0.296 ⫻ 10⫺6

204 ⫻ 107 206 211 214 220 223 227 230 228 225 221 216 207 ⫻ 107

7.62 ⫻ 10⫺2 7.54 7.48 7.41 7.36 7.18 7.01 6.82 6.68 6.50 6.30 6.12 5.94 ⫻ 10⫺2

0.610 0.872 1.13 1.60 2.34 4.24 7.38 12.3 19.9 31.2 47.3 70.1 101.3

Properties of Air at Atmospheric Pressure Temperature T(°C) ⫺30 ⫺20 ⫺10 0 10 20 30 40 50 60 70 80 90 100 200 300

Density r(kg/m3)

Viscosity m(N ⭈ s/m2)

Kinematic viscosity n (m2/s)

1.452 1.394 1.342 1.292 1.247 1.204 1.164 1.127 1.092 1.060 1.030 1.000 0.973 0.946 0.746 0.616

1.56 ⫻ 10⫺5 1.61 1.67 1.72 1.76 1.81 1.86 1.91 1.95 2.00 2.05 2.09 2.13 2.17 2.57 2.93 ⫻ 10⫺5

1.08 ⫻ 10⫺5 1.16 1.24 1.33 1.42 1.51 1.60 1.69 1.79 1.89 1.99 2.09 2.19 2.30 3.45 4.75 ⫻ 10⫺5

Speed of sound c(m/s) 312 319 325 331 337 343 349 355 360 366 371 377 382 387 436 480



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Mechanics of Fluids, Fourth Edition Merle C. Potter, David C. Wiggert, Bassem Ramadan, and Tom I-P. Shih Publisher, Global Engineering: Christopher M. Shortt Senior Acquisitions Editor: Randall Adams Senior Developmental Editor: Hilda Gowans

Page iv

© 2012, 2002 Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.

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Library of Congress Control Number: on file ISBN-13: 978-0-495-66773-5 ISBN-10: 0-495-66773-0

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Printed in the United States of America 1 2 3 4 5 6 7 13 12 11 10



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Mechanics of Fluids Fourth Edition Merle C. Potter Michigan State University

David C. Wiggert Michigan State University

Bassem Ramadan Kettering University with

Tom I-P. Shih Purdue University

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States



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Contents CHAPTER 1 BASIC CONSIDERATIONS 3 1.1 Introduction 4 1.2 Dimensions, Units, and Physical Quantities 4 1.3 Continuum View of Gases and Liquids 8 1.4 Pressure and Temperature Scales 11 1.5 Fluid Properties 14 1.6 Conservation Laws 23 1.7 Thermodynamic Properties and Relationships 24 1.8 Summary 30 Problems 32 CHAPTER 2 FLUID STATICS 39 2.1 Introduction 40 2.2 Pressure at a Point 40 2.3 Pressure Variation 41 2.4 Fluids at Rest 43 2.5 Linearly Accelerating Containers 67 2.6 Rotating Containers 69 2.7 Summary 72 Problems 74 CHAPTER 3 INTRODUCTION TO FLUIDS IN MOTION 3.1 Introduction 88 3.2 Description of Fluid Motion 88 3.3 Classification of Fluid Flows 100 3.4 The Bernoulli Equation 107 3.5 Summary 116 Problems 117

87

CHAPTER 4 THE INTEGRAL FORMS OF THE FUNDAMENTAL LAWS 127 4.1 Introduction 128 4.2 The Three Basic Laws 128 4.3 System-to-Control-Volume Transformation 132 v



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Contents

4.4 4.5 4.6 4.7 4.8

Conservation of Mass 137 Energy Equation 144 Momentum Equation 157 Moment-of-Momentum Equation 176 Summary 179 Problems 182

CHAPTER 5 THE DIFFERENTIAL FORMS OF THE FUNDAMENTAL LAWS 203 5.1 Introduction 204 5.2 Differential Continuity Equation 205 5.3 Differential Momentum Equation 210 5.4 Differential Energy Equation 223 5.5 Summary 229 Problems 231 CHAPTER 6 DIMENSIONAL ANALYSIS AND SIMILITUDE 6.1 Introduction 238 6.2 Dimensional Analysis 239 6.3 Similitude 248 6.4 Normalized Differential Equations 258 6.5 Summary 262 Problems 263

237

CHAPTER 7 INTERNAL FLOWS 271 7.1 Introduction 272 7.2 Entrance Flow and Developed Flow 272 7.3 Laminar Flow in a Pipe 274 7.4 Laminar Flow between Parallel Plates 281 7.5 Laminar Flow between Rotating Cylinders 288 7.6 Turbulent Flow in a Pipe 292 7.7 Uniform Turbulent Flow in Open Channels 325 7.8 Summary 329 Problems 331 CHAPTER 8 EXTERNAL FLOWS 345 8.1 Introduction 346 8.2 Separation 350 8.3 Flow Around Immersed Bodies 352 8.4 Lift and Drag on Airfoils 367 8.5 Potential-Flow Theory 372 8.6 Boundary-Layer Theory 385 8.7 Summary 409 Problems 411



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Contents vii

CHAPTER 9 COMPRESSIBLE FLOW 425 9.1 Introduction 426 9.2 Speed of Sound and the Mach Number 427 9.3 Isentropic Nozzle Flow 431 9.4 Normal Shock Wave 442 9.5 Shock Waves in Converging-Diverging Nozzles 449 9.6 Vapor Flow through a Nozzle 454 9.7 Oblique Shock Wave 456 9.8 Isentropic Expansion Waves 461 9.9 Summary 465 Problems 466

CHAPTER 10 FLOW IN OPEN CHANNELS 473 10.1 Introduction 474 10.2 Open-Channel Flows 475 10.3 Uniform Flow 478 10.4 Energy Concepts 484 10.5 Momentum Concepts 498 10.6 Nonuniform Gradually Varied Flow 510 10.7 Numerical Analysis of Water Surface Profiles 518 10.8 Summary 528 Problems 529

CHAPTER 11 FLOWS IN PIPING SYSTEMS 543 11.1 Introduction 544 11.2 Losses in Piping Systems 544 11.3 Simple Pipe Systems 550 11.4 Analysis of Pipe Networks 561 11.5 Unsteady Flow in Pipelines 574 11.6 Summary 582 Problems 583

CHAPTER 12 TURBOMACHINERY 599 12.1 Introduction 600 12.2 Turbopumps 600 12.3 Dimensional Analysis and Similitude for Turbomachinery 617 12.4 Use of Turbopumps in Piping Systems 626 12.5 Turbines 632 12.6 Summary 647 Problems 648



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viii Contents

CHAPTER 13 MEASUREMENTS IN FLUID MECHANICS 13.1 Introduction 656 13.2 Measurement of Local Flow Parameters 656 13.3 Flow Rate Measurement 664 13.4 Flow Visualization 673 13.5 Data Acquisition and Analysis 681 13.6 Summary 693 Problems 693

655

CHAPTER 14 COMPUTATIONAL FLUID DYNAMICS 697 14.1 Introduction 698 14.2 Examples of Finite-Difference Methods 699 14.3 Stability, Convergence, and Error 710 14.4 Solution of Couette Flow 717 14.5 Solution of Two-Dimensional Steady-State Potential Flow 721 14.6 Summary 726 References 728 Problems 729 APPENDIX 733 A. Units and Conversions and Vector Relationships 733 B. Fluid Properties 735 C. Properties of Areas and Volumes 741 D. Compressible-Flow Tables for Air 742 E. Numerical Solutions for Chapter 10 751 F. Numerical Solutions for Chapter 11 758 BIBLIOGRAPHY 773 References 773 General Interest 774 ANSWERS TO SELECTED PROBLEMS INDEX

776

785



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Preface The motivation to write a book is difficult to describe. Most often the authors suggest that the other texts on the subject have certain deficiencies that they will correct, such as an accurate description of entrance flows and flows around blunt objects, the difference between a one-dimensional flow and a uniform flow, the proper presentation of the control volume derivation, or a definition of laminar flow that makes sense. New authors, of course, introduce other deficiencies that future authors hope to correct! And life goes on. This is another fluids book that has been written in hopes of presenting an improved view of fluid mechanics so that the undergraduate can understand the physical concepts and follow the mathematics. This is not an easy task: Fluid mechanics is a subject that contains many difficult-to-understand phenomena. For example, how would you explain the hole scooped out in the sand by the water on the upstream side of an abutment? Or the high concentration of smog contained in the Los Angeles area (it doesn’t exist to the same level in New York)? Or the unexpected strong wind around the corner of a tall building in Chicago? Or the vibration and subsequent collapse of a large concrete-steel bridge due to the wind? Or the trailing vortices observed from a large aircraft? We have attempted to present fluid mechanics so that the student can understand and analyze many of the important phenomena encountered by the engineer. The mathematical level of this book is based on previous mathematics courses required in all engineering curricula. We use solutions to differential equations and vector algebra. Some use is made of vector calculus with the use of the gradient operator, but this is kept to a minimum since it tends to obscure the physics involved. Many popular texts in fluid mechanics have not presented fluid flows as fields. That is, they have presented primarily those flows that can be approximated as one-dimensional flows and have treated other flows using experimental data. We must recognize that when a fluid flows around an object, such as a building or an abutment, its velocity possesses all three components which depend on all three space variables and often, time. If we present the equations that describe such a general flow, the equations are referred to as field equations, and velocity and pressure fields become of interest. This is quite analogous to electrical and ix



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Preface

magnetic fields in electrical engineering. In order for the difficult problems of the future, such as large-scale environmental pollution, to be analyzed by engineers, it is imperative that we understand fluid fields. Thus in Chapter 5 we introduce the field equations and discuss several solutions for some relatively simple geometries. The more conventional manner of treating the flows individually is provided as an alternate route for those who wish this more standard approach. The field equations can then be included in a subsequent course. Perhaps a listing of the additions made in this fourth edition would be of interest. We have: • Included inside the back cover the second edition of a DVD entitled “Multimedia Fluid Mechanics,” by G.M. Homsy, distributed by Cambridge University Press. A number of virtual labs and demonstrations have been identified at various places in the text. • Added many examples and problems, most of which have real-life applications. • Collected all the multiple-choice problems to the front of the problem sets. They can be used to review the subject of Fluid Mechanics for the Fundamentals of Engineering and the GRE/Engineering exams. • Deleted the chapter on Environmental Fluid Mechanics in an attempt to shorten the book. • Simplified the chapter on Computational Fluid Mechanics. • Made numerous changes to clarify the presentation. The introductory material included in Chapters 1 through 9 has been selected carefully to introduce students to all fundamental areas of fluid mechanics. Not all of the material in each chapter need be covered in an introductory course. The instructor can fit the material to a selected course outline. Some sections at the end of each chapter may be omitted without loss of continuity in later chapters. In fact, Chapter 5 can be omitted in its entirety if it is decided to exclude field equations in the introductory course, a relatively common decision. That chapter can then be included in an intermediate fluid mechanics course. After the introductory material has been presented, there is sufficient material to present in one or two additional courses. This additional course or courses could include material that had been omitted in the introductory course and combinations of material from the more specialized Chapters 9 through 14. Much of the material is of interest to all engineers, although several of the chapters are of interest only to particular disciplines. We have included examples worked out in detail to illustrate each important concept presented in the text material. Numerous home problems, many having multiple parts for better homework assignments, provide the student with ample opportunity to gain experience solving problems of various levels of difficulty. Answers to selected home problems are presented just prior to the Index. We have also included design-type problems in several of the chapters. After studying the material, reviewing the examples, and working several of the home problems, students should gain the needed capability to work many of the problems encountered in actual engineering situations. Of course, there are numerous classes of problems that are extremely difficult to solve, even for an experienced



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Preface xi

engineer. To solve these more difficult problems, the engineer must gain considerably more information than is included in this introductory text. There are, however, many problems that can be solved successfully using the material and concepts presented herein. Many students take the FE/EIT exam at the end of their senior year, the first step in becoming a professional engineer. The problems in the FE/EIT exam are all four-part, multiple choice. Consequently, we have included this type of problem at the beginning of the chapters. Multiple-choice problems will be presented using SI units since the FE/EIT exam uses SI units exclusively. Additional information on the FE/EIT exam can be obtained from a website at ppi2pass.com. The book is written emphasizing SI units; however, all properties and dimensional constants are given in English units as well. Approximately one-fifth of the examples and problems are presented using English units. The authors are very much indebted to both their former professors and to their present colleagues. Chapter 10 was written with inspiration from F.M. Henderson’s book titled Open Channel Flow (1996), and D. Wood of the University of Kentucky encouraged us to incorporate comprehensive material on pipe network analysis in Chapter 11. Several illustrations in Chapter 11 relating to the water hammer phenomenon were provided by C.S. Martin of the Georgia Institute of Technology. R.D. Thorley provided some of the problems at the end of Chapter 12. Tom Shih assisted in the writing of Chapter 14 on Computational Fluid Dynamics. Thanks to Richard Prevost for writing the MATLAB® solutions. We would also like to thank our reviewers: Sajjed Ahmed, University of Nevada; Mohamed Alawardy, Louisiana State University; John R. Biddle, California State Polytechnic University; Nancy Ma, North Carolina State University; Saeed Moaveni, Minnesota State University; Nikos J. Mourtos, San Jose (CA) State University; Julia Muccino, Arizona State University; Emmanuel U. Nzewi, North Carolina A&T State University; and Yiannis Ventikos, Swiss Federal Institute of Technology. Merle C. Potter David C. Wiggert Bassem Ramadan



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Multimedia Fluid Mechanics The following entries from the Multimedia Fluid Mechanics DVD have been added to the text.They are experiments and demonstrations that are included on the DVD inside the back cover. The instructor may assign additional files from the DVD at appropriate locations in the text. Also, the reader may find selected entries that are not listed of particular interest. To view a file on a specified page, simply open any of the eight major headings, then enter a page number in the box at the top and click on “go to page.” The page numbers after the descriptors refer to the pages on the DVD. Example 1.4a: Capillary Rise, 512 Example 4.7a: Pipe Flow Virtual Lab, 947–948 Example 4.11a: Pipe Elbow Example, 918–923 Example 4.12a: Fire Hose Example, 911–917 Example 4.19a: Pressure-Jet Virtual Lab, 932–935 Example 6.2a: Oscillations in a U-Tube, 547–550 Example 6.2b: Flow from a Tank, 555–558 Example 6.2c: Geometric and Dynamic Similarity, 542–543 Example 7.5a: Taylor Cells, page 24 Example 7.8a: Turbulent Boundary-Layer Lab, 858–860 Example 8.1a: Forces on an Airfoil Example, 924–931 Example 8.2a: Cylinder-Wake Virtual Lab, 936–938 Example 8.3a: Example of Vortex Shedding, 195–197 Example 8.12a: Potential-Flow Virtual Lab, page 295 Example 8.13a: Viscous Layer Growth, 619–621 Example 8.14a: Profiles in a Turbulent Plume, 861–864 xii



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Multimedia Fluid Mechanics xiii

Example 8.17a: Laminar Boundary-Layer Growth, 625–627 Problem 8.14: Flow Past a Sphere, 5 Below Fig. 3.13: Taylor Cells, 24 In margin, p.105: Reynolds Number, 524 In margin, p.105: Pipe Flow, 202 In margin, p.380: A Doublet, 281 In margin, p.239: Dimensional Analysis, 588 In margin, p.242: Dimensional Analysis, 521–523 In margin, p.246: Dimensionless Numbers, 524–528 In margin, p.249: Dynamic Similarity, 195 In margin, p.250: Similarity and Scaling, 494, 534, 535, 568 In margin, p.274: Laminar Flow in a Pipe, 686 In margin, p.288: Flow between Cylinders, 735 In margin, p.293: Reynolds Decompostion, 689 In margin, p.295: Turbulent Flow in a Pipe, 687 In margin, p.347: Flow Past a Cylinder, 116, 131, 190 In margin, p.347: Flow over an Airfoil, 649 In margin, p.349: Separated Flow over an Airfoil, 167 In margin, p.350: Separated Flow past Sharp Edges, 662, 664, 666 In margin, p.353: Flow around Immersed Bodies, 652, 657, 694 In margin, p.355: Drag Curve for a Golf Ball, 265 In margin, p.359: Vortex Shedding, 81, 216 In margin, p.362: Streamlining, 651 In margin, p.370: Trailing Vortex, 725, 577 In margin, p.373: Potential Flows, 123, 271 In margin, p.378: Simple Potential Flows, 277 In margin, p.387: Boundary Layers, 163, 260, 602, 622, 677, 852 In margin, p.426: Compressible Flows, 50 In margin, p.347: Speed of Sound, 57 In margin, p.754: CFD Solutions, 669–672, 807, 821



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Nomenclature for quick reference A - area A2, A3 - profile type a - acceleration, speed of a pressure wave a - acceleration vector ax, ay, az - acceleration components B - bulk modulus of elasticity, free surface width b - channel bottom width C - centroid, Chezy coefficient, Hazen-Williams coefficient C1, C3 - profile type CD - drag coefficient Cd - discharge coefficient Cf - skin-friction coefficient CH - head coefficient CL - lift coefficient CP - pressure recovery factor, pressure coefficient CNPSH - net positive suction head coefficient CQ - flow-rate coefficient CV - velocity coefficient CW˙ - power coefficient c - specific heat, speed of sound, chord length, celerity cf - local skin-friction coefficient cp - constant pressure specific heat c√ - constant volume specific heat c.s. - control surface c.v. - control volume D - diameter D ᎏᎏ - substantial derivative Dt d - diameter dx - differential distance du - differential angle E - energy, specific energy, coefficient Ec - critical energy EGL - energy grade line Eu - Euler number e - the exponential, specific energy, wall roughness height, pipe wall thickness xiv



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Nomenclature xv

exp - the exponential e F - force vector F - force FB - buoyant force FH - horizontal force component FV - vertical force component FW - body force equal to the weight f - friction factor, frequency G - center of gravity G 苶M 苶 - metacentric height g - gravity vector g - gravity H - enthalpy, height, total energy H2, H3 - profile type HD - design head HP - pump head HT - turbine head HGL - hydraulic grade line h - distance, height, specific enthalpy hj - head loss across a hydraulic jump I - second moment of an area I苶 - second moment about the centroidal axis Ixy - product of inertia î - unit vector in the x-direction jˆ - unit vector in the y-direction kˆ - unit vector in the z-direction K - thermal conductivity, flow coefficient Kc - contraction coefficient Ke - expansion coefficient Ku√ - correlation coefficient k - ratio of specific heats L - length LE - entrance length Le - equivalent length ᐉ - length ᐉm - mixing length M - molar mass, Mach number, momentum function M - Mach number M1, M2, M3 - profile type m - mass, side-wall slope, constant for curve fit ˙ - mass flux m ˙ r - relative mass flux m ma - added mass m1, m2 - side-wall slopes ˙ - momentum flux mom N - general extensive property, an integer, number of jets NPSH - net positive suction head n - normal direction, number of moles, power-law exponent, Manning number nˆ - unit normal vector



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Nomenclature

P - power, force, wetted perimeter p - pressure Q - flow rate (discharge), heat transfer QD - design discharge Q˙ - rate of heat transfer q - source strength, specific discharge, heat flux R - radius, gas constant, hydraulic radius, radius of curvature Re - Reynolds number Recrit - critical Reynolds number Ru - universal gas constant Rx, Ry - force components r - radius, coordinate variable r - position vector S - specific gravity, entropy, distance, slope of chennel, slope of EGL S1, S2, S3 - profile type Sc - critical slope St - Strouhal number S - position vector S0 - slope of channel bottom s - specific entropy, streamline coordinate sˆ - unit vector tangent to streamline sys - system T - temperature, torque, tension t - time, tangential direction U - average velocity U앝 - free-stream velocity away from a body u - x-component velocity, circumferential blade speed u⬘ - velocity perturbation u˜ - specific internal energy u 苶 - time average velocity ut - shear velocity V - velocity Vc - critical velocity Vss - steady-state velocity V - velocity vector V - spatial average velocity 苶 V - volume VB - blade velocity Vn - normal component of velocity Vr - relative speed Vt - tangential velocity n - velocity, y-component velocity n⬘ - velocity perturbation nr, nz, nu, nf - velocity components W - work, weight, change in hydraulic grade line ˙ - work rate (power) W ˙ f - actual power W We - Weber number ˙ S - shaft work (power) W



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Nomenclature xvii

v - z-component velocity, velocity of a hydraulic bore XT - distance where transition begins x - coordinate variable xm - origin of moving reference frame x˜ - distance relative to a moving reference frame x苶 - x-coordinate of centroid Y - upstream water height above top of wier y - coordinate variable, flow energy head yp - distance to center of pressure y苶 - y-coordinate of centroid yc - critical depth z - coordinate variable a - angle, angle of attack, lapse rate, thermal diffusivity, kinetic-energy correction factor, blade angle b - angle, momentum correction factor, fixed jet angle, blade angle ⌬ - a small increment ⵜ - gradient operator ⵜ2 - Laplacian d - boundary layer thickness d(x) - Dirac-delta function dd - displacement thickness dn - viscous wall layer thickness 6 - a small volume 6xx, 6xy, 6xz - rate-of-strain components f - angle, coordinate variable, velocity potential function, speed factor G - circulation, vortex strength g - specific weight h - a general intensive property, eddy viscosity, efficiency, a position variable hP - pump efficiency hT - turbine efficiency l - mean free path, a constant, wave length m - viscosity, doublet strength n - kinematic viscosity p - a pi term u - angle, momentum thickness, laser beam angle r - density V - angular velocity VP - specific speed of a pump VT - specific speed of a turbine V - angular velocity vector s - surface tension, cavitation number, circumferential stress sxx, syy, szz - normal stress components t - stress vector t苶 - time average stress txy, txz, tyz - shear stress components v - angular velocity, vorticity ␻ - vorticity vector c - stream function ⭸ ᎏᎏ - partial derivative ⭸x Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.




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Left: Contemporary windmills are used to generate electricity at many locations in the United States. They are located in areas where consistent prevailing winds exist. (IRC/Shutterstock) Top right: Hurricane Bonnie, Atlantic Ocean about 800 km from Bermuda. At this stage in its development, the storm has a well-developed center, or “eye,” where air currents are relatively calm. Vortex-like motion occurs away from the eye. (U.S. National Aeronautics and Space Administration) Bottom right: The Space Shuttle Discovery leaves the Kennedy Space Center on October 29, 1998. In 6 seconds the vehicle cleared the launch tower with a speed of 160 km/h, and in about two minutes it was 250 km down range from the Space Center, 47 km above the ocean with a speed of 6150 km/h. The wings and rudder on the tail are necessary for successful reentry as it enters the earth’s atmosphere upon completion of its mission. (U.S. National Aeronautics and Space Administration)


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1st Pass Pages

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1 Basic Considerations Outline 1.1 1.2 1.3 1.4 1.5

Introduction Dimensions, Units, and Physical Quantities Continuum View of Gases and Liquids Pressure and Temperature Scales Fluid Properties 1.5.1 Density and Specific Weight 1.5.2 Viscosity 1.5.3 Compressibility 1.5.4 Surface Tension 1.5.5 Vapor Pressure 1.6 Conservation Laws 1.7 Thermodynamic Properties and Relationships 1.7.1 Properties of an Ideal Gas 1.7.2 First Law of Thermodynamics 1.7.3 Other Thermodynamic Quantities 1.8 Summary

Chapter Objectives The objectives of this chapter are to: ▲ ▲ ▲ ▲

Introduce many of the quantities encountered in fluid mechanics including their dimensions and units. Identify the liquids to be considered in this text. Introduce the fluid properties of interest. Present the thermodynamic laws and associated quantities.

3


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Chapter 1 / Basic Considerations

1.1 INTRODUCTION

KEY CONCEPT We will present the fundamentals of fluids so that engineers are able to understand the role that fluid plays in particular applications.

A proper understanding of fluid mechanics is extremely important in many areas of engineering. In biomechanics the flow of blood and cerebral fluid are of particular interest; in meteorology and ocean engineering an understanding of the motions of air movements and ocean currents requires a knowledge of the mechanics of fluids, chemical engineers must understand fluid mechanics to design the many different kinds of chemical-processing equipment; aeronautical engineers use their knowledge of fluids to maximize lift and minimize drag on aircraft and to design fan-jet engines; mechanical engineers design pumps, turbines, internal combustion engines, air compressors, air-conditioning equipment, pollution-control equipment, and power plants using a proper understanding of fluid mechanics; and civil engineers must also utilize the results obtained from a study of the mechanics of fluids to understand the transport of river sediment and erosion, the pollution of the air and water, and to design piping systems, sewage treatment plants, irrigation channels, flood control systems, dams, and domed athletic stadiums. It is not possible to present fluid mechanics in such a way that all of the foregoing subjects can be treated specifically; it is possible, however, to present the fundamentals of the mechanics of fluids so that engineers are able to understand the role that the fluid plays in a particular application. This role may involve the proper sizing of a pump (the horsepower and flow rate) or the calculation of a force acting on a structure. In this book we present the general equations, both integral and differential, that result from the conservation of mass principle, Newton’s second law, and the first law of thermodynamics. From these a number of particular situations will be considered that are of special interest. After studying this book the engineer should be able to apply the basic principles of the mechanics of fluids to new and different situations. In this chapter topics are presented that are directly or indirectly relevant to all subsequent chapters. We include a macroscopic description of fluids, fluid properties, physical laws dominating fluid mechanics, and a summary of units and dimensions of important physical quantities. Before we can discuss quantities of interest, we must present the units and dimensions that will be used in our study of fluid mechanics.

1.2 DIMENSIONS, UNITS, AND PHYSICAL QUANTITIES Before we begin the more detailed studies of fluid mechanics, let us discuss the dimensions and units that will be used throughout the book. Physical quantities require quantitative descriptions when solving an engineering problem. Density is one such physical quantity. It is a measure of the mass contained in a unit volume. Density does not, however, represent a fundamental dimension. There are nine quantities that are considered to be fundamental dimensions: length, mass, time, temperature, amount of a substance, electric current, luminous intensity, plane angle, and solid angle. The dimensions of all other quantities can be expressed in terms of the fundamental dimensions. For example, the quantity


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Sec. 1.2 / Dimensions, Units, and Physical Quantities 5

“force” can be related to the fundamental dimensions of mass, length, and time. To do this, we use Newton’s second law, named after Sir Isaac Newton (1642–1727), expressed in simplified form in one direction as F ma

(1.2.1)

Using brackets to denote “the dimension of,” this is written dimensionally as [F ] [m][a] L F M T2

(1.2.2)

where F, M, L, and T are the dimensions of force, mass, length, and time, respectively. If force had been selected as a fundamental dimension rather than mass, a common alternative, mass would have dimensions of [F ] [m] [a] FT 2 M L

(1.2.3)

where F is the dimension1 of force. There are also systems of dimensions in which both mass and force are selected as fundamental dimensions. In such systems conversion factors, such as a gravitational constant, are required; we do not consider these types of systems in this book, so they will not be discussed. To give the dimensions of a quantity a numerical value, a set of units must be selected. In the United States, two primary systems of units are presently being used, the British Gravitational System, which we will refer to as English units, and the International System, which is referred to as SI (Système International) units. SI units are preferred and are used internationally; the United States is the only major country not requiring the use of SI units, but there is now a program of conversion in most industries to the predominant use of SI units. Following this trend, we have used primarily SI units. However, as English units are still in use, some examples and problems are presented in these units as well. The fundamental dimensions and their units are presented in Table 1.1; some derived units appropriate to fluid mechanics are given in Table 1.2. Other units that are acceptable are the hectare (ha), which is 10 000 m2, used for large areas; the metric ton (t), which is 1000 kg, used for large masses; and the liter (L), which is 0.001 m3. Also, density is occasionally expressed as grams per liter (g/L). In chemical calculations the mole is often a more convenient unit than the kilogram. In some cases it is also useful in fluid mechanics. For gases the

KEY CONCEPT SI units are preferred and are used internationally.

1

Unfortunately, the quantity force F and the dimension of force [F ] use the same symbol.


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Chapter 1 / Basic Considerations Table 1.1 Fundamental Dimensions and Their Units Quantity

Dimensions

Length ᐉ Mass m Time t Electric current i Temperature T Amount of substance Luminous intensity Plane angle Solid angle

L M T

SI units meter kilogram second ampere kelvin kg-mole candela radian steradian

M

English units m kg s A K kmol cd rad sr

foot slug second ampere Rankine lb-mole candela radian steradian

ft slug sec A °R lbmol cd rad sr

kilogram-mole (kmol) is the quantity that fills the same volume as 32 kilograms of oxygen at the same temperature and pressure. The mass (in kilograms) of a gas filling that volume is equal to the molecular weight of the gas; for example, the mass of 1 kmol of nitrogen is 28 kilograms. When expressing a quantity with a numerical value and a unit, prefixes have been defined so that the numerical value may be between 0.1 and 1000. These Table 1.2 Derived Units Quantity

Dimensions

Area A Volume V

L2 L3

Velocity V Acceleration a Angular velocity v Force F

L/T L/T 2 T 1 ML/T 2

Density r Specific weight g Frequency f Pressure p

M/L3 M/L2T 2 T1 M/LT 2

Stress t

M/LT 2

Surface tension s Work W

M/T 2 ML2/T 2

Energy E

ML2/T 2

Heat rate Q˙ Torque T Power P ˙ W Viscosity m Mass flux m ˙ Flow rate Q Specific heat c Conductivity K

ML2/T 3 ML2/T 2 ML2/T 3 M/LT M/T L3/T L2/T 2 ML/T 3

SI units m2 m3 L (liter) m/s m/s2 rad/s kgm/s2 N (newton) kg/m3 N/m3 s1 N/m2 Pa (pascal) N/m2 Pa (pascal) N/m Nm J (joule) Nm J (joule) J/s Nm J/s W (watt) Ns/m2 kg/s m3/s J/kgK W/mK

English units ft2 ft3 ft/sec ft/sec2 rad/sec slug-ft/sec2 lb (pound) slug/ft3 lb/ft3 sec1 lb/ft2 (psf) lb/ft2 (psf) lb/ft ft-lb ft-lb Btu/sec ft-lb ft-lb/sec lb-sec/ft2 slug/sec ft3/sec Btu/slug-°R lb/sec-°R


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Sec. 1.2 / Dimensions, Units, and Physical Quantities 7 Table 1.3 SI Prefixes Multiplication factor

Prefix

Symbol

1012 109 106 103 102 103 106 109 1012

tera giga mega kilo centia milli micro nano pico

T G M k c m n p

a

Permissible if used alone as cm, cm2, or cm3.

prefixes are presented in Table 1.3. Using scientific notation, however, we use powers of 10 rather than prefixes (e.g., 2 106 N rather than 2 MN). If larger numbers are written the comma is not used; twenty thousand would be written as 20 000 with a space and no comma.2 Newton’s second law relates a net force acting on a rigid body to its mass and acceleration. This is expressed as F ma

KEY CONCEPT When using SI units, if larger numbers are written (5 digits or more), the comma is not used. The comma is replaced by a space (i.e., 20 000).

(1.2.4)

Consequently, the force needed to accelerate a mass of 1 kilogram at 1 meter per second squared in the direction of the net force is 1 newton; using English units, the force needed to accelerate a mass of 1 slug at 1 foot per second squared in the direction of the net force is 1 pound. This allows us to relate the units by

N kg m/s2

lb slug-ft/sec2

(1.2.5)

which are included in Table 1.2. These relationships between units are often used in the conversion of units. In the SI system, weight is always expressed in newtons, never in kilograms. In the English system, mass is usually expressed in slugs, although pounds are used in some thermodynamic relations. To relate weight to mass, we use W mg

KEY CONCEPT The relationship N kgm/s 2 is often used in the conversion of units.

(1.2.6)

where g is the local gravity. The standard value for gravity is 9.80665 m/s2 (32.174 ft/sec2) and it varies from a minimum of 9.77 m/s2 at the top of Mt. Everest to a maximum of 9.83 m/s2 in the deepest ocean trench. A nominal value of 9.81 m/s2 (32.2 ft/sec2) will be used unless otherwise stated. Finally, a note on significant figures. In engineering calculations we often do not have confidence in a calculation beyond three significant digits since the information 2

In many countries commas represent decimal points, so they are not used where confusion may occur.


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KEY CONCEPT We will assume that all information given is known to three significant digits.

given in the problem statement is often not known to more than three significant digits; in fact, viscosity and other fluid properties may not be known to even three significant digits. The diameter of a pipe may be stated as 2 cm; this would, in general, not be as precise as 2.000 cm would imply. If information used in the solution of a problem is known to only two significant digits, it is incorrect to express a result to more than two significant digits. In the examples and problems we will assume that all information given is known to three significant digits, and the results will be expressed accordingly. If the numeral 1 begins a number, it is not counted in the number of significant digits, i.e., the number 1.210 has three significant digits.

Example 1.1 A mass of 100 kg is acted on by a 400-N force acting vertically upward and a 600-N force acting upward at a 45° angle. Calculate the vertical component of the acceleration. The local acceleration of gravity is 9.81 m/s2. Solution The first step in solving a problem involving forces is to draw a free-body diagram with all forces acting on it, as shown in Fig. E1.1. y 600 N 45°

W

400 N

Fig. E1.1 Next, apply Newton’s second law (Eq. 1.2.4). It relates the net force acting on a mass to the acceleration and is expressed as Fy may Using the appropriate components in the y-direction, with W mg, we have 400 600 sin 45° 100 9.81 100ay ay 1.567 m/s2 The negative sign indicates that the acceleration is in the negative y-direction, i.e., down. Note: We have used only three significant digits in the answer since the information given in the problem is assumed known to three significant digits. (The number 1.567 has three significant digits. A leading “1” is not counted as a significant digit.)

1.3 CONTINUUM VIEW OF GASES AND LIQUIDS Substances referred to as fluids may be liquids or gases. In our study of the fluid mechanics we restrict the liquids that are studied. Before we state the restriction,


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Sec. 1.3 / Continuum View of Gases and Liquids 9

we must define a shearing stress. A force F that acts on an area A can be decomposed into a normal component Fn and a tangential component Ft, as shown in Fig. 1.1. The force divided by the area upon which it acts is called a stress. The force vector divided by the area is a stress vector3, the normal component of force divided by the area is a normal stress, and the tangential force divided by the area is a shear stress. In this discussion we are interested in the shear stress t. Mathematically, it is defined as Ft t lim A 씮 0 A

(1.3.1)

Our restricted family of fluids may now be identified; the fluids considered in this book are those liquids and gases that move under the action of a shear stress, no matter how small that shear stress may be. This means that even a very small shear stress results in motion in the fluid. Gases obviously fall within this category of fluids, as do water and tar. Some substances, such as plastics and catsup, may resist small shear stresses without moving; a study of these substances is included in the subject of rheology and is not included in this book. It is worthwhile to consider the microscopic behavior of fluids in more detail. Consider the molecules of a gas in a container. These molecules are not stationary but move about in space with very high velocities. They collide with each other and strike the walls of the container in which they are confined, giving rise to the pressure exerted by the gas. If the volume of the container is increased while the temperature is maintained constant, the number of molecules impacting on a given area is decreased, and as a result, the pressure decreases. If the temperature of a gas in a given volume increases (i.e., the velocities of the molecules increase), the pressure increases due to increased molecular activity. Molecular forces in liquids are relatively high, as can be inferred from the following example. The pressure necessary to compress 20 grams of water vapor at 20 °C into 20 cm3, assuming that no molecular forces exist, can be shown by the ideal gas law to be approximately 1340 times the atmospheric pressure. Of course, this pressure is not required, because 20 g of water occupies 20 cm3. It follows that the cohesive forces in the liquid phase must be very large. Despite the high molecular attractive forces in a liquid, some of the molecules at the surface escape into the space above. If the liquid is contained, an equilibrium is established between outgoing and incoming molecules. The presence of molecules above the liquid surface leads to a so-called vapor pressure. n

ΔF

3

Normal stress: The normal component of force divided by the area. Shear stress: The tangential force divided by the area. Liquid: A state of matter in which the molecules are relatively free to change their positions with respect to each other but restricted by cohesive forces so as to maintain a relatively fixed volume.4 Gas: A state of matter in which the molecules are practically unrestricted by cohesive forces. A gas has neither definite shape nor volume.

KEY CONCEPT Fluids considered in this text are those that move under the action of a shear stress, no matter how small that stress may be.

ΔF n Components ΔA

ΔA

Fig. 1.1

Stress vector: The force vector divided by the area.

ΔF t

Normal and tangential components of a force.

A quantity that is defined in the margin is in bold face whereas a quantity not defined in the margin is italic.

4

Handbook of Chemistry and Physics, 40th ed. CRC Press, Boca Raton, Fla.


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Continuum: Continuous distribution of a liquid or gas throughout a region of interest.

This pressure increases with temperature. For water at 20°C this pressure is approximately 0.02 times the atmospheric pressure. In our study of fluid mechanics it is convenient to assume that both gases and liquids are continuously distributed throughout a region of interest, that is, the fluid is treated as a continuum. The primary property used to determine if the continuum assumption is appropriate is the density r, defined by m r lim v씮0 V

Standard atmospheric conditions: A pressure of 101.3 kPa and temperature of 15°C.

KEY CONCEPT To determine if the continuum model is acceptable, compare a length l with the mean free path. Mean free path: The average distance a molecule travels before it collides with another molecule.

(1.3.2)

where m is the incremental mass contained in the incremental volume V. The density for air at standard atmospheric conditions, that is, at a pressure of 101.3 kPa (14.7 psi) and a temperature of 15°C (59°F), is 1.23 kg/m3 (0.00238 slug/ft3). For water, the nominal value of density is 1000 kg/m3 (1.94 slug/ft3). Physically, we cannot let V씮0 since, as V gets extremely small, the mass contained in V would vary discontinuously depending on the number of molecules in V; this is shown graphically in Fig. 1.2. Actually, the zero in the definition of density should be replaced by some small volume , below which the continuum assumption fails. For most engineering applications, the small volume shown in Fig. 1.2 is extremely small. For example, there are 2.7 1016 molecules contained in a cubic millimeter of air at standard conditions; hence, is much smaller than a cubic millimeter. An appropriate way to determine if the continuum model is acceptable is to compare a characteristic length l (e.g., the diameter of a rocket) of the device or object of interest with the mean free path l, the average distance a molecule travels before it collides with another molecule; if l >> l, the continuum model is acceptable. The mean free path is derived in molecular theory. It is m l 0.225 2 rd

(1.3.3)

where m is the mass (kg) of a molecule, r the density (kg/m3) and d the diameter (m) of a molecule. For air m 4.8 1026 kg and d 3.7 1010 m. At standard atmospheric conditions the mean free path is approximately 6.4 106 cm, ρ

ε

ΔV

Fig. 1.2 Density at a point in a continuum.


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Sec. 1.4 / Pressure and Temperature Scales

11

at an elevation of 100 km it is 10 cm, and at 160 km it is 5000 cm. Obviously, at higher elevations the continuum assumption is not acceptable and the theory of rarefied gas dynamics (or free molecular flow) must be utilized. Satellites are able to orbit the earth if the primary dimension of the satellite is of the same order of magnitude as the mean free path. With the continuum assumption, fluid properties can be assumed to apply uniformly at all points in a region at any particular instant in time. For example, the density r can be defined at all points in the fluid; it may vary from point to point and from instant to instant; that is, in Cartesian coordinates r is a continuous function of x, y, z, and t, written as r (x, y, z, t).

1.4 PRESSURE AND TEMPERATURE SCALES In fluid mechanics pressure results from a normal compressive force acting on an area. The pressure p is defined as (see Fig. 1.3) F p lim n A씮0 A

(1.4.1)

where Fn is the incremental normal compressive force acting on the incremental area A. The metric units to be used on pressure are newtons per square meter (N/m2) or pascal (Pa). Since the pascal is a very small unit of pressure, it is more conventional to express pressure in units of kilopascal (kPa). For example, standard atmospheric pressure at sea level is 101.3 kPa. The English units for pressure are pounds per square inch (psi) or pounds per square foot (psf). Atmospheric pressure is often expressed as inches of mercury or feet of water, as shown in Fig. 1.4; such a column of fluid creates the pressure at the bottom of the column, providing the column is open to atmospheric pressure at the top. Both pressure and temperature are physical quantities that can be measured using different scales. There exist absolute scales for pressure and temperature, and there are scales that measure these quantities relative to selected reference points. In many thermodynamic relationships (see Section 1.7) absolute scales must be used for pressure and temperature. Figures 1.4 and 1.5 summarize the commonly used scales. The absolute pressure reaches zero when an ideal vacuum is achieved, that is, when no molecules are left in a space; consequently, a negative absolute pressure is an impossibility. A second scale is defined by measuring pressures relative

KEY CONCEPT In many relationships, absolute scales must be used for pressure and temperature. Absolute pressure: The scale measuring pressure, where zero is reached when an ideal vacuum is achieved.

ΔF n

Surface ΔA

Fig. 1.3 Definition of pressure.


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Chapter 1 / Basic Considerations A

A – Positive pressure

pA gage

B – Negative pressure or positive vacuum

Standard atmosphere

Local atmosphere p gage (negative) B

pA absolute

101.3 kPa 14.7 psi 2117 psf 30.0 in. Hg 760 mm Hg 34 ft H2O 1.013 bar

p = 0 gage

B p absolute B p = 0 absolute

Zero absolute pressure

Fig. 1.4 Gage pressure and absolute pressure. Gage pressure: The scale measuring pressure relative to the local atmospheric pressure.

to the local atmospheric pressure. This pressure is called gage pressure. A conversion from gage pressure to absolute pressure can be carried out using pabsolute patmospheric pgage

KEY CONCEPT Whenever the absolute pressure is less than the atmospheric pressure, it may be called a vacuum.

Vacuum: Whenever the absolute pressure is less than the atmospheric pressure.

(1.4.2)

Note that the atmospheric pressure in Eq. 1.4.2 is the local atmospheric pressure, which may change with time, particularly when a weather “front” moves through. However, if the local atmospheric pressure is not given, we use the value given for a particular elevation, as given in Table B.3 of Appendix B, and assume zero elevation if the elevation is unknown. The gage pressure is negative whenever the absolute pressure is less than atmospheric pressure; it may then be called a vacuum. In this book the word “absolute” will generally follow the pressure value if the pressure is given as an absolute pressure (e.g., p 50 kPa absolute). If it were stated as p 50 kPa, the pressure would be taken as a gage pressure, except that atmospheric pressure is always an absolute pressure. Most often in fluid mechanics gage pressure is used.

Steam point

°C

K

°F

°R

100°

373

212°

672°

0°

273

32°

492°

–18°

255

0°

460°

Ice point Special point

Zero absolute temperature

Fig. 1.5 Temperature scales.


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Sec. 1.4 / Pressure and Temperature Scales

13

Two temperature scales are commonly used, the Celsius (C) and Fahrenheit (F) scales. Both scales are based on the ice point and steam point of water at an atmospheric pressure of 101.3 kPa (14.7 psi). Figure 1.5 shows that the ice and steam point are 0 and 100°C on the Celsius scale and 32 and 212°F on the Fahrenheit scale. There are two corresponding absolute temperature scales. The absolute scale corresponding to the Celsius scale is the kelvin (K) scale. The relation between these scales is K °C 273.15

(1.4.3)

The absolute scale corresponding to the Fahrenheit scale is the Rankine scale (°R). The relation between these scales is °R °F 459.67

(1.4.4)

Note that in the SI system we do not write 100°K but simply 100 K, which is read “100 kelvins,” similar to other units. Reference will often be made to “standard atmospheric conditions” or “standard temperature and pressure.” This refers to sea-level conditions at 40° latitude, which are taken to be 101.3 kPa (14.7 psi) for pressure and 15°C (59°F) for temperature. Actually, The standard pressure is usually taken as 100 kPa, sufficiently accurate for engineering calculations.

KEY CONCEPT In the SI system, we write 100 K, which is read “100 kelvins.”

Example 1.2 A pressure gage attached to a rigid tank measures a vacuum of 42 kPa inside the tank shown in Fig. E1.2, which is situated at a site in Colorado where the elevation is 2000 m. Determine the absolute pressure inside the tank.

air

–42 kPa

Fig. E1.2

Solution To determine the absolute pressure, the atmospheric pressure must be known. If the elevation were not given, we would assume a standard atmospheric pressure of 100 kPa. However, with the elevation given, the atmospheric pressure is found from Table B.3 in Appendix B to be 79.5 kPa. Thus p 42 79.5 37.5 kPa absolute Note: A vacuum is always a negative gage pressure. Also, using standard atmospheric pressure of 100 kPa is acceptable, rather than 101.3 kPa, since it is within 1%, which is acceptable engineering accuracy.


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1.5 FLUID PROPERTIES In this section we present several of the more common fluid properties. If density variation or heat transfer is significant, several additional properties, not presented here, become important.

1.5.1 Density and Specific Weight Specific weight: Weight per unit volume (g rg).

Fluid density was defined in Eq. 1.3.2 as mass per unit volume. A fluid property directly related to density is the specific weight g or weight per unit volume. It is defined by

g

Specific gravity: The ratio of the density of a substance to the density of water.

mg W rg V V

(1.5.1)

where g is the local gravity. The units of specific weight are N/m3 (lb/ft3). For water we use the nominal value of 9800 N/m3 (62.4 lb/ft3). The specific gravity S is often used to determine the specific weight or density of a fluid (usually a liquid). It is defined as the ratio of the density of a substance to the density of water at a reference temperature of 4°C: r g S rwater gwater

KEY CONCEPT Specific gravity is often used to determine the density of a fluid.

(1.5.2)

For example, the specific gravity of mercury is 13.6, a dimensionless number; that is, the mass of mercury is 13.6 times that of water for the same volume. The density, specific weight, and specific gravity of air and water at standard conditions are given in Table 1.4. The density and specific weight of water do vary slightly with temperature; the approximate relationships are (T 4)2 rH2O 1000 180

(1.5.3)

(T 4)2 gH2O 9800 18 TABLE 1.4 Density, Specific Weight, and Specific Gravity of Air and Water at Standard Conditions Density r slug/ft

1.23 1000

0.0024 1.94

kg/m Air Water

Specific weight g

3

3

N/m3

lb/ft3

Specific gravity S

12.1 9810

0.077 62.4

0.00123 1


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Sec. 1.5 / Fluid Properties

15

For mercury the specific gravity relates to temperature by SHg 13.6 0.0024T

(1.5.4)

Temperature in the three equations above is measured in degrees Celsius. For temperatures under 50°C, using the nominal values stated earlier for water and mercury, the error is less than 1%, certainly within engineering limits for most design problems. Note that the density of water at 0°C (32°F) is less than that at 4°C; consequently, the lighter water at 0°C rises to the top of a lake so that ice forms on the surface. For most other liquids the density at freezing is greater than the density just above freezing.

1.5.2 Viscosity Viscosity can be thought of as the internal stickiness of a fluid. It is one of the properties that influences the power needed to move an airfoil through the atmosphere. It accounts for the energy losses associated with the transport of fluids in ducts, channels, and pipes. Further, viscosity plays a primary role in the generation of turbulence. Needless to say, viscosity is an extremely important fluid property in our study of fluid flows. The rate of deformation of a fluid is directly linked to the viscosity of the fluid. For a given stress, a highly viscous fluid deforms at a slower rate than a fluid with a low viscosity. Consider the flow of Fig. 1.6 in which the fluid particles move in the x-direction at different speeds, so that particle velocities u vary with the ycoordinate. Two particle positions are shown at different times; observe how the particles move relative to one another. For such a simple flow field, in which u u(y), we can define the viscosity m of the fluid by the relationship du t m dy

Viscosity: The internal stickiness of a fluid.

KEY CONCEPT Viscosity plays a primary role in the generation of turbulence.

(1.5.5)

where t is the shear stress of Eq. 1.3.1 and u is the velocity in the x-direction. The units of t are N/m2 or Pa (lb/ft2), and of m are Ns/m2 (lb-sec/ft2).The quantity du/dy is a velocity gradient and can be interpreted as a strain rate. Stress velocity-gradient relationships for more complicated flow situations are presented in Chapter 5. The concept of viscosity and velocity gradients can also be illustrated by considering a fluid within the small gap between two concentric cylinders, as shown

Strain rate: The rate at which a fluid element deforms.

y X t=0

t = t1 t = 2t1 t = 3t1

u(y) Particle 1 Particle 2

Fig. 1.6

X X

Relative movement of two fluid particles in the presence of shear stresses.


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Chapter 1 / Basic Considerations L u

h

R

ω

(a)

r

(b)

u

τ T

R

Rω u (d) r=R

r=R+h

r

(c)

Fig. 1.7 Fluid being sheared between cylinders with a small gap: (a) the two cylinders; (b) rotating inner cylinder; (c) velocity distribution; (d) the inner cylinder. The outer cylinder is fixed and the inner cylinder is rotating.

in Fig. 1.7. A torque is necessary to rotate the inner cylinder at constant rotational speed v while the outer cylinder remains stationary. This resistance to the rotation of the cylinder is due to viscosity. The only stress that exists to resist the applied torque for this simple flow is a shear stress, which is observed to depend directly on the velocity gradient; that is,

冨冨

du t m dr

(1.5.6)

where du/dr is the velocity gradient and u is the tangential velocity component, which depends only on r. For a small gap (h 0 (an unfavorable gradient)

y

y

y

U(x)

∂u/∂y (d) dp/dx > 0 (separated flow)

u

Fig. 8.28

∂ 2 u/∂y 2

Influence of the pressure gradient.

If a positive (unfavorable) pressure gradient is imposed on the flow, the second derivative at the wall will be positive and the flow will be as sketched in part (c) or (d). If the unfavorable pressure gradient acts over a sufficient distance, part (d) will probably represent the flow situation with the flow separated from the surface. Near the wall the higher pressure downstream will drive the low momentum flow near the wall in the upstream direction, resulting in flow reversal, as shown. The point at which u/y 0 at the wall locates the point of separation.


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Sec. 8.7 / Summary 409

The problem of a laminar boundary layer with a pressure gradient can be solved using conventional numerical techniques. The procedure is relatively simple using the simplified boundary-layer equation (8.6.45) with a known pressure gradient. For a turbulent flow the Reynolds stress term must be included; research continues to develop models of turbulent quantities that will result in acceptable numerical solutions. Experimental results are often necessary for turbulent-flow problems, as was the situation for internal flows.

8.7 SUMMARY The drag and lift coefficients are defined as Drag CD 2 1 rV A 2

Lift CL 2 1 rV A 2

(8.7.1)

where the area is the projected area for blunt objects, and the chord times the length for an airfoil. Vortex shedding occurs from a cylinder whenever the Reynolds number is in the range 300 Re 10 000. The frequency of shedding is found from the Strouhal number fD St V

(8.7.2)

where f is the frequency, in hertz. Plane potential flows are constructed by superimposing the following simple flows: c U y q Line source: c 2pu Irrotational vortex: c ln r 2p m Doublet: c sin u r Uniform flow:

(8.7.3)

The stream function for the rotating cylinder is given by m G ccylinder U y sin u ln r r 2p

(8.7.4)

where the cylinder radius is rc

m 冪莦 U

(8.7.5)


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Chapter 8 / External Flows

The velocity components are c u y

c v x

1 c vr r u

(8.7.6)

c vu r

For a laminar boundary layer on a flat plate with zero pressure gradient the exact solution provides d5

n 冪莦 U x

cf 0.664

n 冪莦 xU

Cf 1.33

n 冪莦 LU

(8.7.7)

For a turbulent flow from the leading edge, the power-law profile with h 7 gives

冢冣

n 0.38x xU

1/5

冢冣

xU cf 0.059 n

1/5

冢

n Cf 0.073 LU

冣

1/5

(8.7.8)

where the wall shear and drag force per unit width are, respectively, 1 t0 cf rU 2 2

1 FD Cf rU 2L 2

(8.7.9)

FUNDAMENTALS OF ENGINEERING EXAM REVIEW PROBLEMS 8.1

The drag force on a streamlined shape is due primarily to: (A) The wake (B) The component of the pressure force acting in the flow direction (C) The shear stress (D) The separated region near the trailing edge

8.2

A golf ball has dimples to increase its flight distance. Select the best reason that accounts for the longer flight distance of a dimpled ball compared to that of a smooth ball (A) The shearing stress is smaller on the dimpled ball (B) The dimpled ball has an effective smaller diameter (C) The wake on the dimpled ball is smaller (D) The pressure on the front of the smooth ball is larger

8.3

Flood waters at 10°C flow over an 8-mm diameter fence wire with a velocity of 0.8 m/s. Which of the following is true? (A) It is a Stokes flow with no separation (B) The separated region covers most of the rear of the wire (C) The drag is due to the relatively low pressure in the separated region (D) The separated region covers only a small area at the rear of the wire

8.4

The drag on a 10-m-diameter spherical water storage tank in an 80 km/h wind is approximately: (A) 6300 N (C) 3200 N (B) 4700 N (D) 2300 N

8.5

A 4-m-long smooth cylinder experiences a drag of 60 N when subjected to an atmospheric air speed of 40 m/s. Estimate the diameter of the cylinder. (A) 127 mm (C) 26 mm (B) 63 mm (D) 4.1 mm


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Problems 411 8.6

8.7

Vortices are shed from a 2-cm-diameter cylinder due to a 4 m/s air stream. How far apart would you expect the vortices to be downstream of the cylinder? (A) 44 cm (C) 9 cm (B) 23 cm (D) 4 cm Streamlining reduces drag primarily by: (A) Reducing the wall shear (B) Reducing the pressure in the stagnation region (C) Reducing the separated flow area (D) Eliminating the wake

8.8

Estimate the takeoff speed needed for a 1200-kg aircraft (including payload) if the angle of attack at takeoff is to be 10°. The effective wing area (chord times length) is 16 m2. (A) 22 m/s (B) 33 m/s (C) 44 m/s (D) 55 m/s

PROBLEMS Separated Flows 8.9

8.10

8.11

8.12

8.13

8.14

Sketch the flow over an airfoil at a large angle of attack for both attached flow and separated flow. Also, sketch the expected pressure distributions on the top and bottom surfaces for both flows. Identify regions of favorable and unfavorable pressure gradients. A spherical particle is moving in 20°C atmospheric air at a velocity of 20 m/s. What must its diameter be for Re 5 and Re 105? Sketch the expected flow field at these Reynolds numbers. Identify all regions of the flow. Sketch the flow that is expected over a semitruck (a tractor and trailer) where the trailer is substantially higher than the tractor with and without an air deflector attached to the roof of the tractor. Sketch a side view indicating any separated regions, boundary layers, and the wake. Air is blowing by a long, rectangular building with the wind blowing parallel to the long sides. Sketch the top view showing the separated flow regions, the inviscid flow region, the boundary layers, and the wake. A 0.8-in.-diameter sphere is to move with Re 5. At what velocity does it travel if it is submerged in: (a) Water at 60°F? (b) Water at 180°F? (c) Standard air at 60°F? Air at 20°C flows around a cylindrical body at a velocity of 20 m/s. Calculate the Reynolds number if the body is: (a) A 6-m-diameter smokestack (b) A 6-cm-diameter flagpole (c) A 6-mm-diameter wire Use Re VD/n. Would a separated flow be expected?

8.15

The pressure distribution on the front of a 2-m-diameter disk (Fig. P8.15) is approximated by p(r) p0(1 r2). If V 20 m/s in this 20°C–atmospheric airflow, estimate the drag force and the drag coefficient for this disk. Assume that the pressure on the back side is zero.

p(r) r

V

Fig. P8.15 8.16

A flat plate 30 cm 30 cm acts as a hydrofoil. If it is oriented at an angle of attack of 10°, estimate the lift and the drag if the pressure on the bottom side is 20 kPa and on the top side there is a vacuum of 10 kPa; neglect the effect of shear stress. Also, estimate the lift and drag coefficients if the hydrofoil velocity is 5 m/s. Use the surface area of the plate in the definition of the lift and drag coefficients.

8.17

The symmetric airfoil shown in Fig. P8.17 flies at an altitude of 12 000 m with an angle of attack of 5°. If pl 26 kPa and pu 8 kPa, estimate the lift and drag coefficients neglecting shear stresses. 5°

pu

V = 750 m/s Air pl

5°

Fig. P8.17


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8.19

8.20

8.21

8.22

8.23

8.24

8.25

8.26

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Chapter 8 / External Flows If the drag coefficient for a 10-cm-diameter sphere is given by CD 1.0, calculate the drag if the sphere is falling in the atmosphere: (a) At sea level (c) In 10°C water (b) At 30 000 m Calculate the drag on a smooth 50-cm-diameter sphere when subjected to a 20°C–atmospheric airflow of: (a) 6 m/s (b) 15 m/s (c) Flow past a sphere, page 5 A 4.45-cm-diameter golf ball is roughened to reduce the drag during its flight. If the Reynolds number at which the sudden drop occurs is reduced from 3 105 to 6 104 by the roughening (dimples), would you expect this to lengthen the flight of a golf ball significantly? Justify your reasoning with appropriate calculations. A 4-in.-diameter smooth sphere experiences a drag of 0.5 lb when placed in standard 60°F air. (a) What is the velocity of the airstream? (b) At what increased speed will the sphere experience the same drag? A smooth 20-cm-diameter sphere experiences a drag of 4.2 N when placed in a 20°C water channel. Calculate the drag coefficient and the Reynolds number. A 2-m-diameter smokestack stands 60 m high. It is designed to resist a 40-m/s wind. At this speed, what total force would be expected, and what moment would the base be required to resist? Assume atmospheric–20°C air. A flagpole is composed of three sections: a 5-cmdiameter top section that is 10 m long, a 7.5-cm-diameter middle section 15 m long, and a 10-cm-diameter bottom section 20 m long. Calculate the total force acting on the flagpole and the resisting moment provided by the base when subjected to a 25-m/s wind speed. Make the calculations for: (a) A winter day at 30°C (b) A summer day at 35°C A drag force of 10 lb is desired at Re 105 on a 6-ft-long cylinder in a 60°F–atmospheric airflow. What velocity should be selected, and what should be the cylinder diameter? A 20-m-high structure is 2 m in diameter at the top and 8 m in diameter at the bottom as shown in Fig. P8.26. If the diameter varies linearly with height, estimate the total drag force due to a 30-m/s wind. Use atmospheric air at 20°C.

2m 30 m/s

20 m 8m

Fig. P8.26 8.27

8.28

8.29

8.30

8.31

A steel sphere (S 7.82) is dropped in water at 20°C. Calculate the terminal velocity if the diameter of the sphere is: (a) 10 cm (c) 1 cm (b) 5 cm (d) 2 mm Estimate the terminal velocity of a 20-in.-diameter sphere as it falls in a 60°F atmosphere near the Earth if it has a specific gravity of: (a) 0.005 (b) 0.02 (c) 1.0 Estimate the terminal velocity of a skydiver by making reasonable approximations of the arms, legs, head, and body. Assume air at 20°C. Assuming that the drag on a modern automobile at high speeds is due primarily to form drag, estimate the power (horsepower) needed by an automobile with a 3.2-m2 cross-sectional area to travel at: (a) 80 km/h (b) 90 km/h (c) 100 km/h The 2 m 3 m sign shown in Fig. P8.31 weighs 400 N. What wind speed is required to blow over the sign? Sign

V 2m

Thin legs (neglect drag)

20 cm 2.2 m

Fig. P8.31 8.32

8.33

Calculate the drag force on a 60-cm-diameter, 6-m-long cylinder if 20°C-air is blowing normal to its axis at 40 km/h and the cylinder: (a) Is a section of a very long cylinder (b) Has both ends free (c) Is fixed to the ground with the top free. An 80-kg paratrooper leaps from an elevation of 3000 m. Estimate the paratrooper’s landing speed if she: (a) Curls up as tightly as possible (b) Uses an 8-m-diameter lightweight parachute. (c) Uses a 2-m-diameter safety parachute


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Problems 413 8.34

A semitruck travels 200 000 km each year at an average speed of 90 km/h. Estimate the savings if a streamlined deflector is added to reduce the drag coefficient. Fuel costs $0.40 per liter and the truck without the deflector averages 1.2 km per liter of fuel.

8.35

A rectangular car carrier has a cross section 6 ft 2 ft. Estimate the minimum added power (horsepower) required to travel at 60 mph because of the car carrier.

8.36

8.37

8.38

8.39

V 5m

Assume that the velocity at the corners of an automobile where the rear-view mirrors are located is 1.6 times the automobile’s velocity. How much horsepower is required by the two 10-cm-diameter rearview mirrors for an automobile speed of 100 km/h? Atmospheric air at 25°C blows normal to a 4-m-long section of a cone, 30 cm in diameter at one end and 2 m in diameter at the other end, with a velocity of 20 m/s. Predict the drag on the object. Assume that CD 0.4 for each cylindrical element of the object. An 80-cm-diameter balloon (Fig. P8.38) that weighs 0.5 N is filled with 20°C helium at 20 kPa. Neglecting the weight of the string, estimate V if a equals: (a) 80° (b) 70° (c) 60° (d) 50°

2m

60 cm

Fig. P8.39 8.40

8.41

V 4m

For the newly planted tree shown in Fig. P8.39, the soil–ball interface is capable of resisting a moment of 5000 Nm. Predict the minimum wind speed that might possibly tip the tree over. Assume that CD 0.4 for a cylinder in this air flow.

8.42

α

Fig. P8.38

A sign 1.2 m 0.6 m is attached to the top of a pizza delivery car. The car travels 10 hours a day, 6 days a week. Estimate the added cost in a year that the sign adds to the fuel used by the car. The average speed of the car is 40 km/h, fuel costs $0.60 per liter, the engine/drive train is 30% efficient, and the fuel contains 12 000 kJ/kg. A bike rider is able to travel at an average speed of 25 mph while riding upright. It is determined that the rider’s projected area is 0.56 m2. If the rider assumes a racing position so that his projected area is 0.40 m2, estimate his increased speed if his drag coefficient decreases 20%, assuming the same energy expenditure. An automobile with a cross-sectional area of 3 m2 is powered by a 40-hp engine. Estimate the maximum possible speed if the drive train is 90% efficient. (The engine is rated by the power produced before the transmission.)

Vortex Shedding 8.43

8.44

Over what range of velocities would you expect vortex shedding from a telephone wire 3 mm in diameter? Could you hear any of the vorticies being shed? (Human beings can hear frequencies between 20 and 20 000 Hz.) A wire is being towed through 60°F water normal to its axis at a speed of 6 ft/s. What diameter (both large and small) could the wire have so that vortex shedding would not occur?

8.45

8.46

It is quite difficult to measure low velocities. To determine the velocity in a low-speed airflow, the vorticies being shed from a 10-cmdiameter cylinder are observed to occur at 0.2 Hz. Estimate the airspeed if the temperature is 20°C. Motion pictures show that vortices are shed from a 2-m-diameter cylinder at 0.002 Hz while it is moving in 20°C water. What is the cylinder velocity?


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Chapter 8 / External Flows eter steel cable is subjected to a force of 30 000 N. What length cable would result in resonance in a 10-m/s wind? (Note: The third and fifth harmonics may also lead to resonance. Calculate all three lengths.)

The cables supporting a suspended bridge (Fig. P8.47) have a natural frequency of T/(prL2d2) hertz, where T is the tension, r the cable density, d its diameter, and L its length. The vortices shed from the cables may lead to resonance and possible failure. A certain 1.6-cm-diam-

Fig. P8.47

Streamlining 8.48

A 6-in.-diameter smokestack on a semitruck extends 6 ft straight up into the free stream. Estimate the horsepower needed because of the stack for a speed of 60 mph. If the stack were streamlined, estimate the reduced horsepower.

drag on the cylinder. If the cylinder were streamlined, what would be the percentage reduction in drag?

8.49

A wind speed of 3 m/s blows normal to an 8-cmdiameter smooth cylinder that is 2 m long. Calculate the drag force. The cylinder is now streamlined. What is the percentage reduction in the drag? Assume that T 20°C.

8.50

Water flows by an 80-cm-diameter cylinder that protrudes 2 m up from the bottom of a river. For an average water speed of 2 m/s, estimate the

8.51

Circular 2-cm-diameter tubes are used as supports on an ultralight airplane, designed to fly at 50 km/h. If there are 20 linear meters of the tubes, estimate the horsepower needed by the tubes. If the tubes were streamlined, estimate the reduced horsepower required of the tubes.

8.52

A bike racer is able to ride 50 km/h at top speed. Estimate the drag force just due to his head. If he wore a streamlined, tight-fitting helmet, estimate the reduced drag force.

Cavitation 8.53

8.54

The critical cavitation number for a streamlined strut is 0.7. Find the maximum velocity of the body to which the strut is attached if cavitation is to be avoided. The body is traveling 5 m beneath a water surface. A lift force of 200 kN is desired at a speed of 12 m/s on the hydrofoil of Fig. P8.54, designed to operate at a depth of 40 cm. It has a 40-cm chord and is 10 m long. Calculate the angle of attack and the drag force. Is cavitation present under these conditions?

Angle of attack

Fig. P8.54 8.55

A hydrofoil, designed to operate at a depth of 16 in., has a 16-in. chord and is 30 ft long. A lift force of 50,000 lb is desired at a speed of 35 ft/sec. Calculate the angle of attack and the drag force. Is cavitation present under these conditions?


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Problems 415 8.56

A body resembling a sphere has a diameter of approximately 0.8 m. It is towed at a speed of 20 m/s, 5 m below the water surface. Estimate the drag acting on the body.

8.57

A 2200-kg mine sweeper is designed to ride above the water with hydrofoils at the four corners providing the lift. If the hydrofoils have a 40-cm chord length, what total hydrofoil length is required if they are to operate 60 cm below the surface at a 6° angle of attack? The vehicle is to travel at 50 m/s.

Added Mass 8.58

A 40-cm-diameter sphere, which weighs 400 N, is released from rest while submerged in water. Calculate its initial acceleration: (a) Neglecting the added mass (b) Including the added mass

8.59

A submersible, whose length is twice its maximum diameter, resembles an ellipsoid. If its added mass is ignored, what is the percentage error in a calculation of its initial acceleration if its specific gravity is 1.2?

Lift and Drag on Airfoils 8.60

8.61

8.62

8.63

A rectangular wing on a small airplane has a 1.3 m chord and a 10 m span. When flying in air at 220 km/h, the wing experiences a total aerodynamic force of 18 kN. If the lift to drag ratio is 3, determine the lift coefficient of the wing. In a lift-and-drag experiment performed in a wind tunnel, an NACA0012 airfoil was used. The airfoil has a 15cm chord and a span length of 45cm. Using a force measuring device, a lift force of 60 N was measured on the airfoil for a Reynolds number of 4.586 105. The lift coefficient for this airfoil is given by CL 2 sin, where is the angle of attack. Under the given conditions, what is the angle of attack in degrees? An aircraft with a mass, including payload, of 1000 kg is designed to cruise at a speed of 80 m/s at 10 km. The effective wing area is approximately 15 m2. Determine the lift coefficient and the angle of attack. What power is required by the airfoil during cruise? Assume a conventional airfoil. A 1500 kg aircraft is designed to carry a payload of 3000 N when cruising at 80 m/s at a height 10 km. The effective wing area is 20 m2. Assuming a conventional airfoil, calculate: (a) The takeoff speed if an angle of attack of 10° is desired (b) The stall speed when landing (c) The power required at cruise if 45% of the power is needed to move the airfoil

8.64

8.65

8.66

8.67

8.68

In Problem 8.63 it was necessary to assume that takeoff was at sea level. What would be the takeoff speed in Wyoming, where the elevation is 2000 m? The aircraft of Problem 8.63 is flown at 2 km rather than 10 km. Estimate the percentage increase or decrease in power required at cruise. An additional load of 6000 N is added to the aircraft of Problem 8.63. Estimate the takeoff speed if the angle of attack remains at 10°. Estimate the minimum landing speed for a 250 000-kg aircraft in an emergency situation if the angle of attack is selected to be near stall and: (a) No slotted flaps are used (b) One slotted flap is used (c) Two slotted flaps are used Its wingspan is 60 m and the airfoil chord averages 8 m. In Problem 8.67 it is assumed that the aircraft lands at standard sea-level conditions since neither elevation nor temperature is given. Calculate the percentage increase or decrease in the emergency landing speed if the aircraft is to land: (a) In Denver where the elevation is 1600 m (b) At sea level when the temperature is very cold at 40°C (c) At sea level when the temperature is hot at 50°C


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8.69

A proposed aircraft is to resemble a huge airfoil, a flying wing (Fig. P8.69). Its wingspan will be 200 m and its chord will average 30 m. Estimate, assuming a conventional airfoil, the total mass of the aircraft, including payload, for a design speed of 800 km/h at an elevation of 8 km. Also, calculate the power requirement. Fig. P8.69 Vorticity, Velocity Potential, and Stream Function

8.70

8.71

8.72

8.73

8.75

Take the curl of the Navier–Stokes equation and show that the vorticity equation (8.5.3) results. (See Eq. 5.3.20.) Write the three component vorticity equations contained in Eq. 8.5.3 using rectangular coordiˆ nates. Use ␻ vx iˆ vy jˆ vz k. Simplify the vorticity equation 8.5.3 for a plane flow („ 0 and /z 0). Use ␻ (vx, vy, vz). What conclusion can be made about the magnitude of vz in an inviscid, plane flow (such as flow through a short contraction) that contains vorticity? Consider the flow in the contracting channel shown below. It is assumed that a vortex tube (y-vorticity) exists in the boundary layer on the upper and lower plates. Using the vorticity transport equation, explain the existence of x-vorticity downstream of the obstruction.

8.76

8.77

Solid y

Obstruction

8.78 x

Flow In

Vortex tube

Flow Out Solid Top view

z x

8.79

Side view Vortex tubes

8.74

Obstruction

Fig. P8.73 Determine which of the following flows are irrotational and incompressible and find the velocity potential function, should one exist for each incompressible flow. (a) V 10xiˆ 20yjˆ (b) V 8yiˆ 8xjˆ 6zkˆ 2 ˆ 兹x苶 (c) V (xiˆ yj)/ y2苶 (d) V (xiˆ yjˆ )/(x2 y2)

An attempt is to be made to solve Laplace’s equation for flow around a circular cylinder of radius rc oriented in the center of a channel of height 2h. The velocity profile far from the cylinder is uniform. State the necessary boundary conditions. Assume that c 0 at y h. The origin of the coordinate system is located at the center of the cylinder. State the stream function and velocity potential corresponding to a uniform velocity of 100iˆ 50jˆ using rectangular coordinates. A flow is represented by the stream function c 40 tan 1 (y/x). (a) Express the stream function in polar form (b) Is this an incompressible flow? Show why (c) Determine the velocity potential (d) Find the radius where the acceleration is

10 m/s2 The stream function for a flow is c 20 ln (x2 y2) m2/s. Determine the complex velocity potential for this incompressible flow. If the pressure at a large distance from the origin is 20 kPa, what is the pressure at the point (0, 20 cm) if water is flowing? A stream function is given by 10y c 10y x2 y2 (a) Show that this satisfies 2c 0 (b) Find the velocity potential f (x, y) (c) Assuming water to be flowing, find the pressure along the x-axis if p 50 kPa at x (d) Locate any stagnation points.


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Problems 417 8.80

The velocity potential for a flow is

(d) Locate any stagnation points (e) Find the acceleration at x 2 m, y 0

f 10x 5 ln (x2 y2) (a)

Show that this function satisfies Laplace’s equation (b) Find the stream function c (x, y) (c) Assume that water is flowing and find the pressure along the x-axis if p 100 kPa at x

8.81

The velocity profile in a wide 0.2-m-high channel is given by u(y) y y2/0.2. Determine the stream function for this flow. Calculate the flow rate by integrating the velocity profile and by using c. Explain why a velocity potential does not exist by referring to Eq. 8.5.2.

Superposition of Simple Flows 8.82

The body formed by superimposing a source at the origin of strength 5p ft2/sec and a uniform flow of 30 ft/sec is shown in Fig. P8.82. (a) Locate any stagnation points (b) Find the y-intercept yB of the body (c) Find the thickness of the body at x (d) Find u at x 12 in., y 0

8.86

8.87 y

(o,yB) x

U∞ = 30 fps

8.88

Fig. P8.82 8.83

8.84

8.85

A source with strength p m2/s and a sink of equal strength are located at ( 1 m, 0) and (1 m, 0), respectively. They are combined with a uniform flow U 10 m/s to form a Rankine oval. Calculate the length and maximum thickness of the oval. If p 10 kPa at x , find the minimum pressure if water is flowing. An oval is formed from a source and sink of strengths 2p m2/s located at ( 1, 0) and (1, 0), respectively, combined with a uniform flow of 2 m/s. Locate any stagnation points, and find the velocity at ( 4, 0) and (0, 4). Distances are in meters. Two sources of strength 2p m2/s are located at (0, 1) and (0, 1), respectively. Sketch the resulting flow and locate any stagnation points. Find the velocity at (1, 1). Distances are in meters.

8.89

The two sources of Problem 8.85 are superimposed with a uniform flow. Sketch the flow, locate any stagnation points, and find the y-intercept of the body formed if: (a) U 10 m/s (b) U 1 m/s (c) U 0.2 m/s A doublet of strength 60 m3/s is superimposed with a uniform flow of 8 m/s of water. Calculate: (a) The radius of the resulting cylinder (b) The pressure increase from x to the stagnation point (c) The velocity vu (u) on the cylinder. (d) The pressure decrease from the stagnation point to the point of minimum pressure on the cylinder A sink of strength 4p m2/s is superimposed with a vortex of strength 20p m2/s. (a) Sketch a pathline of a particle initially occupying the point (x 0, y 1 m). Use tangents at every 45° with straight-line extensions (b) Calculate the acceleration at (0, 1) (c) If p(10, 10) 20 kPa, what is p (0, 0.1 m) if atmospheric air is flowing? The cylinder shown in Fig. P8.89 is formed by combining a doublet of strength 40 m3/s with a uniform flow of 10 m/s. (a) Sketch the velocity along the y-axis from the cylinder to y (b) Calculate the velocity at (x 4 m, y 3 m) (c) Calculate the drag coefficient for the cylinder assuming potential flow over the front half and constant pressure over the back half


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Chapter 8 / External Flows 8.93

y

8.94

rc

U∞ = 10 m/s

x

8.95

Fig. P8.89 8.90

8.91

8.92

A 2-m-diameter cylinder is placed in a uniform water flow of 4 m/s. (a) Sketch the velocity along the x-axis from the cylinder to x (b) Find vu on the front half of the cylinder (c) Find p(u) on the front half of the cylinder if p 50 kPa at x (d) Estimate the drag force on a 1-m length of the cylinder if the pressure over the rear half is constant and equal to the value at u 90° Superimpose a free stream U 30 ft/sec, a doublet m 400 ft2/sec, and a vortex G 1000 ft2/sec. Locate any stagnation points and predict the minimum and maximum pressure on the surface of the cylinder if p 0 at x and atmospheric air is flowing. A 0.8-m-diameter cylinder is placed in a 20-m/s atmospheric airflow. At what rotational speed should the cylinder rotate so that only one stagnation point exists on its surface? Calculate the minimum pressure acting on the cylinder if p 0 at x .

8.96

A 1.2-m-diameter cylinder rotates at 120 rpm in a 3-m/s atmospheric airstream. Locate any stagnation points and calculate the minimum and maximum pressures on the cylinder if p 0 at x . The circulation around a 60-ft-long airfoil (measured tip to tip) is calculated to have a value of 15,000 ft2/sec. Estimate the lift generated by the airfoil if the aircraft is flying at 30,000 ft with a speed of 350 ft/sec. Assume the flow to be incompressible. The velocity field due to a source of strength 2p m2/s per meter, located at (2 m, 2 m) in a 90° corner, is desired. Use the method of images, that is, add one or more sources at the appropriate locations, and determine the velocity field by finding u (x, y) and v (x, y). Flow through a porous medium is modeled with Laplace’s equation and an associated velocity potential function. Natural gas can be stored in certain underground rock structures for use at a later time. A well is placed next to an impervious rock formation, as shown in Fig. P8.96. If the well is to extract 0.2 m3/s per meter, predict the velocity to be expected at the point (4 m, 3 m). See Problem 8.95 for the method of images.

(6 m, 2 m)

Fig. P8.96

Boundary Layers 8.97

8.98

How far from the leading edge can turbulence be expected on an airfoil traveling at 300 ft/sec if the elevation is: (a) 0 ft? (b) 12,000 ft? (c) 30,000 ft? Use Recrit 6 105 and assume a flat plate with zero pressure gradient. An airfoil on a commercial jetliner can be crudely approximated by a flat plate. Determine the distance from the leading edge at which transition to turbulence can be expected. The jet is traveling at

8.99

800 km/h at an altitude of 9000 m, where the pressure and temperature are 31 kPa and 43°C, respectively. The boundary layer on a flat plate with zero pressure gradient is to be studied in a tunnel transporting 20°C air. How far from the leading edge would you expect to find turbulent flow if U 10 m/s and: (a) The plate is held rigid with high free-stream disturbance level? (b) The plate is held rigid with low free-stream disturbance level?


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Problems 419 (c)

8.100 8.101

8.102

8.103

The plate is vibrated with low free-stream disturbance level? (d) The plate is vibrated with high free-stream disturbance level? (e) At what distance would you expect a small disturbance to grow for the flow of part (b)? Repeat Problem 8.99 but place the flat plate in a channel transporting 20°C water. A laminar region is desired to be at least 2 m long on a smooth rigid flat plate. A wind tunnel and a water channel are available. What maximum speed can be selected for each? Assume low free-stream fluctuation intensity and 20°C for each. Determine the pressure p(x) in the boundary layer and the velocity U(x) at the edge of the boundary layer that would be present on the front of the cylinder of Problem 8.89. Let the pressure at the stagnation point be 20 kPa with water flowing. Let x be measured from the stagnation point; see Figure 8.19. A boundary layer would develop as shown in Fig. P8.103 from the front stagnation point of Problem 8.90. Determine U(x) and p(x) that would be needed to solve for the boundary-layer growth on the cylinder front. Measure x from the stagnation point.

U(x)

Boundary layer x

Fig. P8.103 8.104

Assuming inviscid uniform flow of air through the contraction shown in Figure P8.104, estimate U(x) and dp/dx, which are necessary to solve for the boundary-layer growth on the flat plate. Assume a one-dimensional flow with r 1.0 kg/m3. 2m 10 cm 40 cm

6 m/s

Flat plate

Fig. P8.104 Von Kármán Integral Equation 8.105

Provide the detailed steps to the final form of Eq. 8.6.1 and Eq. 8.6.3. Refer to Figure 8.23.

8.106

Show that the von Kármán integral equation 8.6.4 can be put in the form

dp d t0 d r dx dx

冕

0

d

dU u(U u) dy r dx

冕

冕

d

u dy

0

d

Note that the quantity u dy is only a function of x. 0 8.107 Show that the von Kármán integral equation of Problem 8.106 can be written as d dU t0 r (uU 2) rddU dx dx To accomplish this, we must show that Bernoulli’s equation p rU 2/2 const. can be differentiated to yield dp r dU dU rU dx d dx dx

冕

0

d

U dy

Assume that u U sin(py/2d) in a zero pressure gradient boundary layer. Calculate: (a) d(x) (b) t0(x) (c) v at y d and x 3 m 8.109 Assume a linear velocity profile and find d(x) and t0(x). Compute the percentage error when compared with the exact expressions for a laminar flow. Use dp/dx 0. 8.110 A boundary-layer profile is approximated with: y u 3 U 0 y d/6 d y 1 d/6 y d/2 u U d 3 8.108

冢冢

冣

冣

y 2 u U 3d 3

d/2 y d

Determine d(x) and t0(x); compute the percentage error when compared with the exact expressions for a laminar flow.


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Chapter 8 / External Flows If the walls in a wind-tunnel test section are parallel, the velocity in the center portion of the tunnel will accelerate as shown in Fig. P8.111. To maintain a constant tunnel velocity so that dp/dx 0, show that the walls should be displaced outward a distance dd(x). If a wind tunnel were square, how far should one wall be displaced outward for dp/dx 0?

8.112

The velocity profile at a given x-location in the boundary layer (Fig. P8.112) is assumed to be y y2 u( y) 10 2 2 d d

冢

冣

A streamline is 2 cm from the flat plate at the leading edge. How far is it from the plate when x 3 m (i.e., what is h)? Also, calculate the displacement thickness at x 3 m. Compare the displacement thickness to (h 2) cm.

δ (x) U(x)

δ (x)

Fig. P8.111 10 m/s Streamline

10 m/s

Boundary layer h

3m

u(y)

Fig. P8.112 8.113

It is desired that the test section in a wind tunnel experience a zero pressure gradient. If the test section has a square cross section, what should be the equation of the displacement of one of the walls (three walls will be straight and parallel or perpendicular) if the 30°C–air is pressurized to 160 kPa absolute? Assume a boundary layer profile of u/U 2y앛d y2앛d2. Assume that y 0 at x 0 of the displacement equation y(x). 8.114 Find dd and u for a laminar boundary layer assuming: (a) A cubic profile (b) A parabolic profile (c) That u U sin (py/2d) Compute percentage errors when compared with the exact values of dd 1.72 兹nx앛U 苶 and u 0.644 兹nx앛U 苶

8.115

A laminar flow is maintained in a boundary layer on a 20-ft-long, 15-ft-wide flat plate with 60°F–atmospheric air flowing at 12 ft/sec. Assuming a parabolic profile, calculate: (a) d at x 20 ft (b) t0 at x 20 ft (c) The drag force on one side (d) v at y d and x 10 f 8.116 Work Problem 8.115, but assume a cubic profile.

Laminar and Turbulent Boundary Layers 8.117

Atmospheric air at 20°C flows at 10 m/s over a 2-m-long 4-m-wide flat plate. Calculate the maximum boundary-layer thickness and the drag force on one side assuming: (a) Laminar flow over the entire length (b) Turbulent flow over the entire length

8.118

Fluid flows over a flat plate at 20 m/s. Determine d and t0 at x 6 m if the fluid is: (a) Atmospheric air at 20°C (b) Water at 20°C Neglect the laminar portion


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Problems 421 8.119 Assume a turbulent velocity profile 苶 u U (y/d)1/7. Does this profile satisfy the conditions at y d? Can it give the shear stress at the wall? Plot both a cubic laminar profile and the one-seventh powerlaw profile on the same graph assuming the same boundary-layer thickness. 8.120 Estimate the drag on one side of a 12-ft-long, 51-ft-wide flat plate if atmospheric air at 60°F is flowing with a velocity of 20 ft/sec. Assume that: (a) Recrit 3 105 (b) Recrit 5 105 (c) Recrit 6 105 8.121 A 1-m-long flat plate with a sharp leading edge that is 2 m wide is towed parallel to itself in 20°C water at 1.2 m/s. Estimate the total drag if: (a) Recrit 3 105 (b) Recrit 6 105 (c) Recrit 9 105 8.122 Air moving at 60 km/h is considered to have a zero boundary-layer thickness at a distance of 100 km from shore. At the beach, estimate the boundary-layer thickness and the wall shear using: (a) A one-seventh power law (b) Empirical data Use T 20°C 8.123 For the conditions of Problem 8.122, calculate: (a) The thickness of the viscous wall layer (b) The displacement thickness at the beach 8.124 Atmospheric air at 60°F flows over a flat plate at 300 ft/sec. At x 20 ft, estimate: (a) The local skin friction coefficient (b) The wall shear (c) The viscous wall-layer thickness (d) The boundary-layer thickness

8.125

Water at 20°C flows over a flat plate at 10 m/s. At x 3 m, estimate: (a) The viscous wall layer thickness (b) The velocity at the edge of the viscous wall layer (c) The value of y at the outer edge of the turbulent zone (d) The boundary-layer thickness 8.126 Estimate the total shear drag on a ship traveling at 10 m/s if the sides are flat plates 10 m 100 m with zero pressure gradients. What is the maximum boundary-layer thickness? 8.127 The large 600-m-long, 100-m-diameter, cigarshaped dirigible of Fig. P8.127 is planned for cruises for retired engineers. As a first estimate the drag is calculated assuming it to be a flat plate with zero pressure gradient, with the drag on the nose and rear areas neglected. (a) Estimate the power required of each of four engines if it is to cruise at 15 m/s (b) Estimate the payload if one half of its volume is filled with helium, and its engines, equipment, and structure have a mass of 1.2 106 kg

Fig. P8.127

Laminar Boundary-Layer Equations 8.128 8.129

Assuming that dp/dx 0, show that Eq. 8.6.47 follows from Eq. 8.6.45. Recalling from calculus that c c j c h y j y h y so that 2c (c앛y) j (c앛y) h c 2 y j h y y y y

冢冣

8.130

show that Eq. 8.6.49 follows from Eq. 8.6.47. Show that Eqs. 8.6.51 follow from the preceding equations.

8.131

Solve Eq. 8.6.52 with the appropriate boundary conditions, using a third-order Runge–Kutta scheme (or any other appropriate numerical algorithm) and verify the results of Table 8.5. (This was a Ph.D. thesis project for Blasius before the advent of the computer!) 8.132 A laminar boundary layer exists on a flat plate with atmospheric air at 20°C moving at 5 m/s. At x 2 m, find: (a) The wall shear (b) The boundary-layer thickness (c) The maximum value of v (d) The flow rate through the boundary layer


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8.133

A laminar boundary layer exists on a flat plate with atmospheric air at 60°F moving at 15 ft/ sec. At x 6 ft, find: (a) The wall shear (b) The boundary-layer thickness (c) The maximum value of v (d) The flow rate through the boundary layer 8.134 Water at 20°C flows over a flat plate with zero pressure gradient at 5 m/s. At x 2 m, find: (a) The wall shear (b) The boundary-layer thickness (c) The flow rate through the boundarylayer 8.135 If, when we defined the boundary layer thickness, we defined d to be that y-location where

u 0.999U , estimate the thickness of the boundary layer of: (a) Problem 8.132 (b) Problem 8.133 8.136 Find the y-location where u 0.5U for the boundary layer of Problem 8.132. What is the value of v at that y-location? What is the shear stress there? 8.137 Assume that v remains unchanged between y d and y 10d. Sketch u(y) for 0 y 10d for a flat plate with zero pressure gradient. Now, assume that v 0 at y 10d. Again sketch u(y). Explain by referring to appropriate equations. 8.138 Sketch the boundary-layer velocity profile near the end of the flat plate of Problem 8.115 and show a Blasius profile of the same thickness on the same sketch.

Pressure-Gradient Effects 8.139

Sketch the velocity profiles near and normal to the cylinder’s surface at each of the points indicated in Fig. P8.139. The flow separates at C.

U∞

B C A

8.140

Sketch the expected boundary-layer profiles at each of the points indicated in Fig. P8.140, showing relative thicknesses. The flow undergoes transition to turbulence just after point A. It separates at D. Show all profiles on the same plot. Indicate the sign of the pressure gradient at each point.

D

A

B E

Fig. P8.139

C D

Fig. P8.140





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A new-generation space vehicle is rendered by an artist as it reenters the earth’s atmosphere. A shock wave is shown to be generated at the leading edge of the vessel, signifying that it is traveling faster than the speed of sound. (U.S. National Aeronautics and Space Administration)


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9 Compressible Flow Outline 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

Introduction Speed of Sound and the Mach Number Isentropic Nozzle Flow Normal Shock Wave Shock Waves in Converging-Diverging Nozzles Vapor Flow Through a Nozzle Oblique Shock Wave Isentropic Expansion Waves Summary

Chapter Objectives The objectives of this chapter are to: ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲

Present the various equations needed to solve the uniform-flow problems of a compressible gas flow. Apply the basic equations to isentropic flow through a nozzle. Introduce the normal shock wave. Solve for velocities and pressures when a shock wave exists in a convergingdiverging nozzle. Analyze the supersonic flow of steam through a nozzle. Study the oblique shock wave needed to turn a supersonic flow through an angle. Calculate the angle through which expansion waves can turn a flow around a convex corner. Provide numerous examples and problems that illustrate isentropic flow, flow through normal and oblique shock waves, and the turning of a supersonic flow around a convex corner.

425


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Chapter 9 / Compressible Flow

9.1 INTRODUCTION Compressible flow: The flow of gas in which the density changes significantly between points on a streamline. KEY CONCEPT Not all gas flows are compressible flows.

In this chapter we consider flows of gases in which the density changes significantly between points on a streamline; such flows are called compressible flows. We will consider those problems that can be solved using the integral equations: the continuity equation, the energy equation, and for some problems, the momentum equation. Not all gas flows are compressible flows, neither are all compressible flows gas flows. At low speeds, less than a Mach number (M V/兹kRT 苶 ) of about 0.3, gas flows may be treated as incompressible flows. This is justified because the density variations caused by the flow are insignificant (less than about 3%). Incompressible gas flows occur in a large number of situations of engineering interest; many of these have been considered in earlier chapters. There are many flows, however, in which the density variations must be accounted for. Included among these are airflows around commercial and military aircraft, airflow through jet engines, and the flow of a gas in compressors and turbines. There are examples of compressible effects important in liquid flows; water hammer and compression waves due to underwater blasts are examples of compressible liquid flows. Compressibility of rock accounts for the propagation through the earth’s surface of longitudinal waves due to an earthquake. In this chapter we are concerned with compressibility effects in gas flows. Let us introduce the effects of compressibility into the simplest of flow situations. The velocity at a given streamwise location in a conduit is assumed to be uniform and hence does not vary normal to the flow direction. For this simple uniform flow we recall that the continuity equation takes the form (see Eq. 4.4.5) ˙ r1A1V1 r2A2V2 m

(9.1.1)

The momentum equation for the uniform, compressible flow takes the form (see Eq. 4.6.5) ˙ 2 V1) 兺 F m(V

(9.1.2)

The energy equation, neglecting potential energy changes, is written (see Eqs. 4.5.17 and 4.5.18) ˙ S V 22 V 12 Q˙ W ˙ h2 h1 m 2

(9.1.3)

where we have used h u˜ p/r. Assuming an ideal gas with constant specific heats, the energy equation takes the form ˙ S V 22 V 21 Q˙ W ˙ cp(T2 T1) m 2

(9.1.4)


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Sec. 9.2 / Speed of Sound and the Mach Number 427

or ˙ S V 22 V 21 Q˙ W p2 p1 k ˙ m 2 r1 k 1 r2

冢

冣

(9.1.5)

where we have used the thermodynamic relations h2 h1 cp(T2 T1)

(9.1.6a)

cp R c√

(9.1.6b)

cp k c√

(9.1.6c)

and the ideal-gas law p rRT

(9.1.7)

If we have an interest in calculating the entropy change between two sections, we will use the definition of entropy as S

冕

冨

dQ T

(9.1.8)

reversible

where dQ represents the differential heat transfer. Using the first law, this becomes, for an ideal gas with constant specific heat, T2 p2 s cp ln R ln T1 p1

(9.1.9)

If a process is adiabatic (Q 0) and reversible (no losses), Eq. 9.1.8 shows that the entropy change is zero (i.e., the flow is isentropic). If the flow is isentropic, the relationship above can be used, along with the ideal-gas law and Eqs. 9.1.6b and c, to show that (k 1)/k

冢冣

T2 p2 T1 p1

冢冣

p2 r2 p1 r1

k

(9.1.10)

Note: Temperatures and pressures must be measured on absolute scales. Let us now introduce a parameter, the Mach number, that will be of special interest throughout our study of compressible flow.

9.2 SPEED OF SOUND AND THE MACH NUMBER The speed of sound is the speed at which a pressure disturbance of small amplitude travels through a fluid. It is analogous to the small ripple, a gravity wave, that travels radially outward when a pebble is dropped into a pond. To determine


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Sound wave: It travels with a velocity c relative to a stationary observer.

the speed of sound consider a small pressure disturbance, called a sound wave, to be passing through a pipe, as shown in Fig. 9.1. It travels with a velocity c relative to a stationary observer as shown in part (a); the pressure, density, and temperature will change by the small amounts p, r, and T, respectively. There will also be an induced velocity V in the fluid immediately behind the sound wave. To simplify the problem we will create a steady flow by having the observer travel at the speed of the wave so that the sound wave appears to be stationary. The flow will then approach the wave from the right, with the speed of sound c, as shown in Fig. 9.1b. The flow properties will all change across the wave and the velocity in the downstream flow will be expressed as c V, where V is the small change in velocity. If the flow speed is reduced, the wave would move to the right, and if the flow speed is increased, it would move to the left. For the flow speed equal to the speed of sound, V c, the sound wave would be stationary, as shown. Let us apply the continuity equation and the energy equation to a small control volume enclosing the sound wave, shown in Fig. 9.1c. The continuity equation (9.1.1) takes the form

Speed of sound, 57

rAc (r r)A(c V)

(9.2.1)

rV cr

(9.2.2)

This can be rewritten as

where we have neglected the higher-order term r V; that is, r represents a small percentage change in r so that r r; also, V c. The streamwise-component momentum equation yields pA ( p p)A rAc(c V c)

(9.2.3)

p rc V

(9.2.4)

which simplifies to,

Stationary wave c

Control volume

p + Δp

ρ + Δρ T + ΔT

ΔV

V=0

(a)

p + Δp

p ρ

ρ + Δρ

T

T + ΔT

c + ΔV

c

p ρ

c + ΔV

c

(p + Δp)A

pA

T (b)

(c)

Fig. 9.1 Sound wave: (a) stationary observer; (b) observer moving with the wave; (c) control volume enclosing the wave.


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Sec. 9.2 / Speed of Sound and the Mach Number 429

Combining this with Eq. 9.2.2 results in c

冪莦 r p

(9.2.5)

Since the changes p and r are quite small, we can write p dp ⯝ r dr

(9.2.6)

Small-amplitude, moderate-frequency waves (up to about 18 000 Hz) travel with no change in entropy (isentropically), so that p k const. r

(9.2.7)

dp p k dr r

(9.2.8)

This can be differentiated to give

Using this in Eq. 9.2.5, the speed of sound c is given by

c

冪莦r kp

(9.2.9)

or, using the ideal-gas law, c 兹kRT 苶

(9.2.10)

At high frequency, sound waves generate friction and the process ceases to be isentropic. It is better approximated by an isothermal process. For an ideal gas an isothermal approximation would lead to c 兹RT 苶

(9.2.11)

For small waves traveling through liquids or solids the bulk modulus is used; it has dimensions of pressure and is equal to r dp/dr. For water it has a nominal value of 2110 MPa. It does vary slightly with temperature and pressure. Using Eq. 9.2.5, this leads to a speed of propagation of 1450 m/s for a small-amplitude pressure wave in water. An important quantity used in the study of compressible flow is the dimensionless velocity called the Mach number, introduced in Eq. 3.3.3 as V M c

KEY CONCEPT A smallamplitude pressure wave in water travels 1450 m/s.

(9.2.12)


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KEY CONCEPT The approach of an object moving at M 1 cannot be sensed until it has passed by.

If M 1, the flow is a subsonic flow, and if M 1, it is a supersonic flow. If M 5 it may be referred to as hypersonic flow; such flows will not be considered in this book. If a source of sound waves is at a fixed location, the waves travel radially away from the source with the speed of sound c. Figure 9.2a shows the position of the sound waves after a time increment t and after multiples of t. Part (b) shows a source moving with a velocity V which is less than the speed of sound. Note that the sound waves always propagate ahead of the source so that an airplane traveling at a speed less than the speed of sound will always “announce” its approach. This is not true, however, for an object traveling at a speed greater than the speed of sound, as shown in Fig. 9.2c. The region outside the cone is a zone of silence, so that an approaching object moving at a supersonic speed could not be heard until it passed overhead and the Mach cone, the cone shown, intercepted the observer. From the figure the angle a of the Mach cone is given by c 1 a sin1 sin1 V M

Shock wave: A largeamplitude wave which may be created by a blunt object.

(9.2.13)

The discussion above is limited to small-amplitude sound waves, often called Mach waves. They are formed on the needle nose of an aircraft, or on the leading edge of an airfoil if the leading edge is sufficiently sharp. If the nose is blunt or if the leading edge is not sufficiently sharp, a supersonic aircraft will produce a large-amplitude wave called a shock wave. The shock wave will also form a zone of silence, but the initial angle at the source created by the shock wave will be larger than that of a Mach wave. Shock waves will be considered in subsequent sections.

2cΔt 3cΔt

2cΔt

3cΔt

cΔt

3cΔt

2cΔt

Mach cone

Source

cΔt Source

2α VΔt cΔt

VΔt VΔt (a)

(b)

VΔt

VΔt

VΔt

(c)

Fig. 9.2 Sound waves propagating from a noise source: (a) stationary source; (b) moving source, V c; (c) moving source, V c.


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Sec. 9.3 / Isentropic Nozzle Flow 431

Example 9.1 A needle-nose projectile traveling at a speed with M 3 passes 200 m above the observer of Fig. E9.1. Calculate the projectile’s velocity and determine how far beyond the observer the projectile will first be heard. V

200 m

L

Fig. E9.1 Solution At a Mach number of 3 the velocity is

苶 V Mc M兹kRT 3 21.4 287 J/kg . K 288 K 1021 m앛s where a standard temperature of 15°C has been assumed since no temperature is given. (Remember, kg N . s2/m and J N . m.) Using h as the height and L as the distance beyond the observer (refer to Fig. 9.2c), we have 1 h sin a 兹苶 L2 h2苶 M With the information given,

giving

1 200 2 2 3 2苶兹L 苶 00 L 566 m

Nⴢm Nⴢm m2 Note: The units on kRT are 2 The quantity k is kg ⴢ K K N ⴢ s2/m s. dimensionless.

9.3 ISENTROPIC NOZZLE FLOW There are many applications where gas flows through a section of tube or conduit that has a changing area in which a steady, uniform, isentropic flow is a good approximation to the actual flow situation. The diffuser near the front of a jet engine, exhaust gases passing through the blades of a turbine, the nozzles on a rocket engine, a broken natural gas line, and gas flow measuring devices are all examples of situations that can be modeled with a steady, uniform, isentropic

KEY CONCEPT Steady, uniform, isentropic flow is a good approximation to many flow situations.


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flow. Consider the flow through the infinitesimal control volume shown in Fig. 9.3. With a changing area the continuity equation rAV const.

(9.3.1)

applied between two sections a distance dx apart takes the form rAV (r dr)(A dA)(V dV ) KEY CONCEPT Retain only first-order terms in the differential quantities.

(9.3.2)

Keeping only first-order terms in the differential quantities, Eq. 9.3.2 can be put in the form dV dA dr 0 r V A

(9.3.3)

The energy equation can be written (see Eq. 9.1.5) V2 k p const. 2 k1 r

(9.3.4)

For the present application we have 2 V2 k p (V dV ) k p dp 2 k1 r 2 k 1 r dr

(9.3.5)

or again, retaining only first-order terms, k r dp p dr V dV 0 r2 k1

(9.3.6)

dx

ρ h

ρ + dρ h + dh

T p

T + dT p + dp

V + dV

V

Fig. 9.3

Uniform, isentropic flow.


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For an isentropic process we use Eq. 9.2.8 and there results p V dV k 2 dr 0 r

(9.3.7)

Substituting for dr/r from Eq. 9.3.3, the above becomes

冢

冣

2 dV rV dA 1 V kp A

(9.3.8)

In terms of the speed of sound this is written as

冢

冣

dV V 2 dA 1 V c2 A

(9.3.9)

Introduce the Mach number and we have the very important relationship dV dA (M2 1) V A

(9.3.10)

for an isentropic uniform flow in a changing area. From Eq. 9.3.10 we can make the following observations: 1. If the area is increasing, dA 0, and M 1, we see that dV must be negative, that is, dV 0. The flow is decelerating for this subsonic flow. 2. If the area is increasing and M 1, we see that dV 0; hence the flow is accelerating in the diverging section for this supersonic flow. 3. If the area is decreasing and M 1, then dV 0, resulting in an accelerating flow. 4. If the area is decreasing and M 1, then dV 0, indicating a decelerating flow. 5. At a throat where dA 0, either dV 0 or M 1, or possibly both. If we define a nozzle as a device that accelerates the flow, we see that observations 2 and 3 describe a nozzle and observations 1 and 4 describe a diffuser, a device that decelerates the flow. The supersonic flow leads to rather surprising results: an accelerating flow in an enlarging area and a decelerating flow in a decreasing area. This is, in fact, the situation encountered in a traffic flow on a freeway; hence a supersonic flow might be used to model a traffic flow. Note that the observations above prohibit a supersonic flow in a converging section attached to a reservoir. If a supersonic flow is to be generated by releasing a gas from a reservoir, there must be a converging section in which a subsonic flow accelerates to the throat where M 1 followed by a diverging section in which the flow continues to accelerate with M 1, as shown in Fig. 9.4. This is the type of nozzle observed on rockets used to place satellites into orbit.

Nozzle: A device that accelerates the flow. Diffuser: A device that decelerates the flow.


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Chapter 9 / Compressible Flow Converging section

dA < 0 M 0

dA > 0 M>1 dV > 0

Diverging section

Reservoir T0 p0 V0 = 0

Supersonic flow

C

Vexit

Throat M=1

Fig. 9.4

Supersonic nozzle.

The isentropic flow in the nozzle will now be considered in more detail. The energy equation between the reservoir where V0 0 and any section can be written in the form V2 cpT0 cpT 2 Stagnation quantities: Quantities with a zero subscript at a location where V 0.

(9.3.11)

Quantities with a zero subscript are often called stagnation quantities since they occur at a location where V 0. Recognizing that V Mc, cp/c√ k, cp c√ R, and c 兹kRT 苶, we can write the energy equation as T0 k1 1 M2 T 2

(9.3.12)

For our isentropic flow, the pressure and density ratios are expressed as k앛(k 1)

冢冣 r k1 冢1 M 冣 r 2

p0 k1 1 M2 p 2

1앛(k 1)

0

(9.3.13)

2

If a supersonic flow occurs downstream of the throat, then M 1 at the throat, and denoting this critical area with an asterisk (*) superscript, we have the following critical ratios: T* 2 T0 k1 p* 2 p0 k1 r* 2 r0 k1

冢冢

k앛(k 1)

冣冣

(9.3.14)

1앛(k 1)


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We often make reference to the critical area even though an actual throat does not occur; we can imagine a throat occurring and call it the critical area. In fact, the isentropic flow table, Table D.1 in Appendix D, includes just such an area. For air with k 1.4 the critical values are p* 0.5283 p0

T* 0.8333 T0

r* 0.6340 r0

KEY CONCEPT We reference the critical area even though an actual throat does not occur.

(9.3.15)

We can determine an expression for the mass flux through the nozzle from the equation m˙ rAV p p 苶 AM 兹kRT RT 兹苶 T

冪莦Rk AM

(9.3.16)

Using Eqs. 9.3.12 and 9.3.13, this can be expressed as k앛(1 k) k1 p0 1 M2 2 ˙ 1앛2 m k1 兹苶 T0 1 M2 2

冢

冣

冢

p0

冪莦

k 莦 AM 冪 R 冣

k k1 MA 1 M2 RT0 2

冢

(k 1)앛2(1 k)

冣

(9.3.17)

If we choose the critical area where M* 1, we see that m˙ p0 A*

冪莦冢

k k1 RT0 2

(k 1)앛2(1 k)

冣

(9.3.18)

This shows that the mass flux in the nozzle is only dependent on the reservoir conditions and the critical area A*. By combining Eqs. 9.3.17 and 9.3.18, the area ratio A/A* can be written in terms of the Mach number as

冤

(k 1)앛2(k 1)

冥

A 1 2 (k 1)M2 k1 A* M

(9.3.19)

This ratio is included in Table D.1 for airflow. As a final consideration in our study of isentropic nozzle flow, we will present the influence of reservoir pressure and receiver pressure on the mass flux. First, the converging nozzle will be presented; then the converging–diverging nozzle will follow. The converging nozzle is assumed to be attached to a reservoir, as shown in Fig. 9.5a, with fixed conditions; the pressure in the receiver can be lowered to provide an increasing mass flux through the nozzle, as shown by the left curve in Fig. 9.5b. When the receiver pressure pr reaches the critical pressure


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Chapter 9 / Compressible Flow Choked flow Me = 1

•

m Choked flow Me = 1

pe < pr

Me < 1

pr p0

Ve

T0

pr < pe

pe

ρ0

Increasing p0 (pr const.)

Decreasing pr (p0 const.)

0.5283 p r /p 0

(a)

1.0

1.893 p 0 /p r

(b)

Fig. 9.5

KEY CONCEPT Reducing pr below the critical pressure has no effect on the upstream flow.

KEY CONCEPT If pe pr the flow exiting the nozzle is able to turn sharply.

Converging nozzle.

(for air pr 0.5283 p0), the Mach number Me at the throat (the exit) is unity. As pr is reduced below this critical value the mass flux will not increase, and the condition of choked flow occurs, as shown by the left curve in Fig. 9.5b. If the throat ˙ is only dependent on the is a critical area, that is, M 1 at the throat, then m throat area A* and the reservoir conditions, as indicated by Eq. 9.3.18. Hence, reducing pr below the critical pressure has no effect on the upstream flow. This is reasonable since disturbances travel at the speed of sound; if pr is reduced, the disturbances that would travel upstream, thereby changing conditions, cannot do so since the velocity of the stream at the exit is equal to the speed of sound thereby preventing any disturbances from propagating upstream. If, in the converging nozzle, we keep pr constant and increase the reservoir pressure (keep T0 constant also), a choked flow again occurs when Me 1; however, when p0 is increased still further, we see from Eq. 9.3.18 that the mass flux will increase, as shown by the right curve of Fig. 9.5b. The pressure pe is equal to pr until the Mach number Me is just equal to unity. This begins the condition of choked flow. The exit pressure pe for the choked flow condition will be greater than the receiver pressure pr, a condition that occurs when a gas line ruptures. A note may be in order explaining how it is possible for the flow exit pressure pe to exceed the receiver pressure pr. If pe pr, the flow exiting the nozzle is able to turn rather sharply, causing a flow pattern sketched in Fig. 9.6. This possible flow situation will be studied in Section 9.8. Now, consider the converging–diverging nozzle, shown in Fig. 9.7, with reservoir and receiver as indicated. We will only present the condition of constant reservoir pressure and reduced receiver pressure. For this nozzle we sketch the pressure ratio p/p0 as a function of location in the nozzle for various receiver pressure ratios pr/p0. If pr/p0 1, no flow occurs, corresponding to curve A. If pr is reduced a small amount, curve B results and a subsonic flow exists throughout the nozzle with a minimum pressure occurring at the throat. As the pressure is reduced still further, a pressure is reached that will result in the Mach number at


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pr

pper

pr

pr

Fig. 9.6

Nozzle exit flow for pe pr.

the throat just being unity, as sketched by curve C; the flow remains subsonic throughout, however. Another particular receiver pressure exists, considerably below that of curve C, that will also produce an isentropic flow throughout; it results in curve D. Any receiver pressure in between these two particular pressures will result in a nonisentropic flow in the nozzle; a shock wave, to be studied later, occurs which renders our assumption of isentropic flow invalid. If the receiver pressure is below that associated with curve D, we again find that the nozzle exit pressure pe is greater than the receiver pressure pr. The mass flux in the nozzle increases from curve A to curve C; but as the receiver pressure is reduced below that of curve C, no increase in mass flux can occur since the throat condition will remain unchanged. The mass flux is given by Eq. 9.3.17. Final notes regard nozzle and diffuser effectiveness. The purpose of a nozzle is to convert enthalpy (which can be thought of as stored energy) into kinetic

KEY CONCEPT The purpose of a nozzle is to convert stored energy into kinetic energy.

Throat p0 T0 ρ

Ve pe

0

pr

Reservoir conditions fixed Receiver pressure varies

1.0

A B

1.0

C p/p 0

p r /p 0 D

x

Fig. 9.7

Converging–diverging nozzle.


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˙ S 0. The efficiency hN of a nozzle energy as described by Eq. 9.1.3 with Q˙ W is defined as (KE)actual hN (KE)isentropic h0 he h0 hes

KEY CONCEPT The purpose of a diffuser is to recover the pressure.

where he is the actual exit enthalpy and hes is the isentropic exit enthalpy. Efficiencies are between 90 and 99% with larger nozzles having the higher percentages because the viscous wall effects that account for most of the losses are relatively small with the larger nozzles. The purpose of a diffuser is to slow down the fluid and recover the pressure. For a diffuser we define the pressure recovery factor Cp to be pactual Cp pisentropic

KEY CONCEPT Supersonic nozzles are constructed with large included angles.

(9.3.20)

(9.3.21)

Such factors vary between 40% when the flow actually separates from the wall (the included angle should be less than about 10° for a subsonic section to avoid this separation) to 85%. Vanes can be installed in wide-angled subsonic diffusers such that each passage between vanes expands at an angle of 10° or less. Viscous effects are significantly greater in a diffuser than in a nozzle because of the thicker viscous wall layers. The reader may think that the flow in the supersonic diverging nozzle may also tend to separate from the wall; this is not the case. Expansion fans, a phenomenon provided by nature, to be studied in Section 9.8, allow the supersonic flow to turn rather sharp angles so that supersonic nozzles are constructed with large included angles, such as those attached to the rockets that propel satellites into orbit.

Example 9.2 Air exits from a reservoir maintained at 20°C and 500 kPa absolute into a receiver maintained at (a) 300 kPa absolute and (b) 200 kPa absolute. Estimate the mass flux if the exit area is 10 cm2. Use the equations first and then the isentropic flow table, Table D.1. Refer to Fig. 9.5. Solution To estimate the mass flux we will assume isentropic flow. For air the receiver pressure that would result in Me 1 is pr 0.5283 p0 0.5283 500 264.2 kPa For part (a) Me 1 since pr 264.2 kPa, and for part (b) choked flow occurs and Me 1 since pr 264.2 kPa.


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(a) To find the exit Mach number, Eq. 9.3.13 gives (k 1)앛k

冢冣

p0 k1 1 M2 2 p or p0 2 M 2e k1 p

(k 1)앛k

冢冣

冢冣

2 500 0.4 300

0.2857

2 k1

2 0.7857 0.4

⬖ Me 0.8864

The mass flux is given by Eq. 9.3.17 and is found to be m˙ p0

k k1 MA冢1 M 冣冪莦 RT 2 2

(k 1)앛2(1 k)

0

500 000

1.4 0.4 0.8864 0.001 冢1 0.8864 冣冪莦 287 莦 293 2 2

2.4앛0.8

1.167 kg앛s (b) Choked flow occurs and thus Me 1; Eq. 9.3.18 yields ˙ p0 A* m

k k1 冪莦 RT 冢 2 冣

(k 1)앛2(1 k)

0

500 000 0.001

2.4앛0.8

1.4 2.4 冪莦 287 莦 293 冢 2 冣

1.181 kg앛s

Now, let us use the isentropic flow table (Table D.1) and solve parts (a) and (b). (a) For a pressure ratio of p/p0 300/500 0.6, we interpolate to find 0.6041 0.6 Me 0.02 0.88 0.886 0.6041 0.5913 Te 0.6041 0.6 (0.8606 0.8659) 0.8659 0.864 T0 0.6041 0.5913 Te 0.864 293 253 K The velocity and density are, respectively, V Mc 0.886 兹苶 1.4 苶 287 苶 253 282 m앛s p 300 r 4.13 kg앛m3 RT 0.287 253 The mass flux is then m˙ rAV 4.13 0.001 282 1.165 kg앛s (b) For choked flow we know that Me 1. The table gives (continued)


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Te 0.8333 T0

and

pe 0.5283 p0

Thus the temperature, velocity, and density are, respectively, T 0.8333 293 244.2 K

苶苶 287 苶 244.2 313.2 m앛s V Mc 1 兹1.4 p 0.5283 500 r 3.769 kg앛m3 RT 0.287 244.2 The mass flux is calculated to be m˙ rAV 3.769 0.001 313.2 1.180 kg앛s The results using the equations are essentially the same as those using the tables.

Example 9.3 A converging–diverging nozzle, with an exit area of 40 cm2 and a throat area of 10 cm2, is attached to a reservoir with T 20°C and p 500 kPa absolute. Determine the two exit pressures that result in M 1 at the throat for an isentropic flow. Also, determine the associated exit temperatures and velocities. See Fig. 9.7. Solution The exit pressures we seek are associated with curves C and D of Fig. 9.7. The area ratio is 40 A 4 A* 10 We could solve Eq. 9.3.19 for M using a trial-and-error technique; however, let us use the isentropic flow table, Table D.1. There are two entries for A/A* 4. Interpolation gives p

冢p冣

0 C

p

冢p冣

0 D

4.182 4.0 (0.9823 0.9864) 0.9864 0.9849 4.182 3.673 4.0 3.999 (0.02891 0.02980) 0.02980 0.02979 4.076 3.999

Hence the two exit pressures that will result in isentropic flow are pC 492.4 kPa

and

pD 14.9 kPa

Note the very small pressure difference (7.6 kPa) between receiver and reservoir necessary to create the flow condition of curve C of Fig. 9.7.


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The exit temperature ratios and Mach numbers are interpolated to be

冢TT冣

0.3576(0.9949 0.9961) 0.9961 0.9957

冢TT冣

0.01299(0.3633 0.3665) 0.3665 0.3665

0 C

0 D

MC 0.3576 0.02 0.14 0.147 MD 0.01299 0.02 2.94 2.94 The exit temperatures associated with curves C and D are thus TC 0.9957 293 291.7 K TD 0.3665 293 107.4 K The exit velocities are found from V Mc to be 1.4 苶 287 苶 291.7 50.3 m앛s VC 0.147 兹苶 VD 2.94 兹苶 1.4 苶 287 苶 107.4 611 m앛s

Example 9.4 Gas flows are assumed to be incompressible flows at Mach numbers less than about 0.3. Determine the error involved in calculating the stagnation pressure for an airflow with M 0.3. Solution For an incompressible airflow the energy equation (4.5.20) with no losses would give V 21 p0 p1 r 2 where V0 0 at the stagnation point. The isentropic flow equation (9.3.13), with k 1.4, gives p0 p1(1 0.2M 12)3.5 Use the binomial theorem (1 x)n 1 nx n(n 1)x2/2! as, letting x 0.2M 21,

and express this

p0 p1(1 0.7M 12 0.175M 41 ) This can be written as (see Eqs. 9.2.9 and 9.2.12) p0 p1 p1M 21(0.7 0.175M 21 )


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r1V 12 (0.7 0.175M 12 ) 1.4 V 12 r1 (1 0.25M 21 ) 2 Substituting M1 0.3, we see that V 21 p0 p1 r1 (1 0.0225 ) 2 Comparing this with the incompressible flow equation, we see that the error is only slightly greater than 2%. Hence it is reasonable to approximate a gas flow below M 0.3 (V ⬵ 100 m/s for air at standard conditions) with an incompressible flow for many engineering applications.

9.4 NORMAL SHOCK WAVE

Shock wave: A large disturbance that propagates through a gas.

KEY CONCEPT The changes that occur across a shock wave take place over an extremely short distance.

Small-amplitude disturbances travel at the speed of sound, as was established in Section 9.2. In this section a large-amplitude disturbance will be studied. Its speed of propagation and its effect on other flow properties, such as pressure and temperature, will be considered. Large-amplitude disturbances occur in a number of situations. Examples include the flow in a gun barrel ahead of the projectile, the exit flow from a rocket or jet engine nozzle, the airflow around a supersonic aircraft, and the expanding front due to an explosion. Such large disturbances propagating through a gas are called shock waves. They can be oriented normal to the flow or at oblique angles. In this section we consider only the normal shock wave that occurs in a tube or directly in front of a blunt object; the photograph in Fig. 9.8 shows the shock wave in front of a sphere. The property changes that occur across a shock wave take place over an extremely short distance. For usual conditions the distance is only several mean free paths of the molecules, on the order of 104 mm. Phenomena such as viscous dissipation and heat conduction that occur within the shock wave will not be studied here. We will treat the shock wave as a zero-thickness discontinuity in the flow and allow our integral control volume equations to relate the global quantities of interest. Consider a normal shock wave moving with velocity V1. We can make it stationary by moving the flow in a tube at velocity V1, as shown in Fig. 9.9. The continuity equation, recognizing that A1 A2, is (9.4.1)

r1V1 r 2V2 ˙ S 0, is The energy equation (9.1.5), with Q˙ W V 22 V 12 p2 p1 k 0 2 r1 k 1 r2

冢

冣

(9.4.2)


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Sec. 9.4 / Normal Shock Wave

443

Fig. 9.8 A shock wave is observed in front of a sphere at M 1.53. (Photograph by A. C. Charters. From Album of Fluid Motion, 1982, The Parabolic Press, Stanford, California.)

The momentum equation (9.1.2), with only pressure forces, becomes p1 p2 r1V1(V2 V1)

(9.4.3)

where the areas have divided out. These three equations allow us to determine three unknown quantities; if r1, V1, and p1 are known, we can find r2, V2, p2, and subsequently T2 and M2, using the appropriate equations. It is convenient, however, to express the equations in terms of the Mach numbers M1 and M2. This results in a set of equations that are easier to solve than solving the three simultaneous equations listed above. To do this we write Eq. 9.4.3, using r1V1 r2V2, in the form

冢

冣

冢

冣

r1V 12 r2V 22 p1 1 p2 1 p1 p2

KEY CONCEPT The three equations allow us to determine three unknowns.

(9.4.4)

Control volume

V1 p1

ρ1

V2 p2 ρ 2

Fig. 9.9 Stationary shock wave in a tube.


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Introducing M 2 V 2r/pk (combine Eq. 9.2.9 and 9.2.12), the momentum equation becomes p2 1 kM 21 2 p1 1 kM 2

(9.4.5)

Similarly, the energy equation (9.4.2), with p rRT, is written as

冢

冣

冢

冣

2 2 k 1 V1 k 1 V2 T1 1 T2 1 kRT1 2 kRT2 2

(9.4.6)

or, substituting M2 V 2/kRT: k1 1 M 21 T2 2 T1 k1 1 M 22 2

(9.4.7)

If we substitute r p/RT into the continuity equation (9.4.1), we have p1V1 p2V2 RT1 RT2

(9.4.8)

which becomes, using V M 兹kRT 苶, p2 M2 p1 M1

冪莦T 1 T1

(9.4.9)

2

Substituting for the pressure and temperature ratios from Eqs. 9.4.5 and 9.4.7, the continuity equation takes the form

冢

冣

1앛2

冢

冣

1앛2

k1 k1 M2 1 M 22 M1 1 M 12 2 2 2 1 kM 22 1 kM 1

(9.4.10)

Hence the downstream Mach number is related to the upstream Mach number by

M 22

2 M 21 k1 2k M 21 1 k1

(9.4.11)


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Sec. 9.4 / Normal Shock Wave

445

This allows us to express the pressure and temperature ratios in terms of M1 only. The momentum equation (9.4.5) takes the form p2 2k k1 M 21 p1 k 1 k1

(9.4.12)

and the energy equation becomes

冢

冣冢

冣

k1 2k 1 M 12 M 21 1 T2 2 k1 T1 (k 1)2 M 21 2(k 1)

(9.4.13)

For air, with k 1.4, the preceding three equations reduce to M 21 5 M 22 2 7M 1 1 p2 7M 21 1 p1 6

(9.4.14)

(M 12 5)(7M 12 1) T2 36M 21 T1 From the first of these three equations, we observe: • If M1 1, then M2 1 and no shock wave exists. • If M1 1, then M2 1 and the normal shock wave converts a supersonic flow into a subsonic flow. • If M1 1, then M2 1 and a subsonic flow appears to be converted into a supersonic flow by the presence of a normal shock wave. This possibility is eliminated by the second law since it would demand a decrease in entropy by a process in an isolated system, an impossibility.

KEY CONCEPT A finite wave that converts a subsonic flow into a supersonic flow is an impossibility.

The impossibility stated above is observed by considering the entropy increase, given by T2 p2 s2 s1 cp ln R ln T1 p1 2 (k 1)M 21 1 kM 21 cp ln 2 R ln 2 2 (k 1)M 2 1 kM 2

(9.4.15)


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Chapter 9 / Compressible Flow Δs 0.3 0.2 0.1 1

Fig. 9.10

2

3

M1

Entropy change for a normal shock in air.

For air, with k 1.4, this is plotted in Fig. 9.10, relating M2 to M1, with Eq. 9.4.11. Note the impossible negative entropy change whenever M1 1. The relationship among the thermodynamic properties, listed in the equations above, can be demonstrated graphically with reference to the T–s diagram in Fig. 9.11. The conditions upstream of the normal shock are designated by state 1, and downstream by state 2. Note the dashed line from state 1 to state 2 for the irreversible process that occurs inside the shock wave. The energy equation, with ˙ 0, can be written as Q˙ W S V 12 V 22 T1 T2 2cp 2cp

(9.4.16)

The stagnation temperature is defined as the temperature that would exist if the flow is brought to rest isentropically. Thus the energy equation gives T01 T02

(9.4.17)

as shown in the figure. The substantial decrease in stagnation pressure, p02 p01, is also observed in Fig. 9.11. If the entropy increases from state 1 to state 2, as it must, the stagnation pressure p02 must decrease, as shown, if we are to maintain T01 T02. Gas tables are available that give the pressure ratio, the temperature ratio, the downstream Mach number, and the stagnation pressure ratio as a function of the upstream Mach number. Table D.2 is such a table for k 1.4 and includes the ratios given by Eq. 9.4.14. Note that M2 is always less than unity, p2 is always greater than p1, T2 is always greater than T1, and p02 is always less than p01.


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Sec. 9.4 / Normal Shock Wave T

Upstream stagnation point

p 01

447

Downstream stagnation point p 02

T01 2

V2 –––– 2cp T2 2

p2

2

V1 –––– 2cp p1

T1 1

s1

s2

s

Fig. 9.11 T–s diagram for a normal shock wave.

Example 9.5 A normal shock wave passes through stagnant air at 60°F and atmospheric pressure of 12 psi with a speed of 1500 ft/sec. Calculate the pressure and temperature downstream of the shock wave. Use (a) the equations and (b) the gas tables. p1

V1

p2

V1

V2

Stagnant air Shock wave

Stationary shock wave

Fig. E9.5 Solution We consider the shock wave to be stationary with V1 1500 ft/sec and p1 12 psia. (a) To use the simplified equations (9.4.14) we must know the upstream Mach number. It is V1 V1 M1 c1 兹k苶 RT1

1500 ft/sec 21.4 1716 ft-lb/slug-R 520 R

1.342 (continued)


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The pressure and temperature are then found to be p1(7M 21 1) p2 6 12(7 1.3422 1) 23.21 psia 6 T1(M 21 5)(7M 21 1) T2 36M 21 520(1.3422 5)(7 1.3422 1) 633.1°R 36 1.3422 (b) From part (a), we use M1 1.342. Interpolation in Table D.2 yields p2 1.342 1.34 (1.991 1.928) 1.928 1.934 p1 1.36 1.34 T2 1.342 1.34 (1.229 1.216) 1.216 1.217 T1 1.36 1.34 Using the information given, we have p2 12 1.934 23.21 psia T2 520 1.217 632.8°R

Example 9.6 A normal shock wave propagates through otherwise stagnant air at standard conditions at a speed of 700 m/s. Determine the speed induced in the air immediately behind the shock wave as shown in Fig. E9.6. Use the equations. V V1

Stagnant air

V2

Vinduced

Shock wave

Stationary shock wave

Fig. E9.6 Solution For standard conditions the temperature is 15°C. The upstream Mach number is thus V1 V 700 2.06 M1 1 c1 兹k苶 RT1 兹1.4 苶苶 287 苶 288


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Sec. 9.5 / Shock Waves in Converging–Diverging Nozzles 449

Using Eq. 9.4.14, we find that M 21 5 M2 2 7M 1 1

冢

冣

1앛2

冢

2.062 5 7 2.062 1

冣

1앛2

0.567

T1(M 21 5)(7M 21 1) T2 36M 21 288(2.062 5)(7 2.062 1) 500.2 K 36 2.062 This allows us to calculate V2 M2c2 0.567 兹苶 1.4 苶 287 苶 500.2 254.4 m앛s This velocity assumes a flow with the shock wave stationary and the air approaching the shock wave at 700 m/s. If we superimpose a velocity of 700 m/s moving opposite to V1, we find the induced velocity to be Vinduced V2 V1 254.4 700 446 m앛s where the negative sign means that the induced velocity would be moving to the left if V1 is to the right. The induced velocity would be in the same direction as the propagation of the shock wave. Such large induced velocities are responsible for much of the damage away from the bomb center caused by explosions of high-power bombs.

9.5 SHOCK WAVES IN CONVERGING–DIVERGING NOZZLES The converging–diverging nozzle has already been presented for an isentropic flow; for receiver-to-reservoir pressure ratios between those of curves C and D of Figs. 9.7 and 9.12 (on the next page), shock waves exist in the flow either inside or outside the nozzle. If pr/p0 a (locate a on the vertical axis on the right of Fig. 9.12), a normal shock wave would exist at an internal location in the diverging portion of the nozzle. Usually, the location of the shock wave is prescribed in student problems since a trial-and-error solution is necessary to locate the shock. When pr/p0 b the normal shock wave is located in the exit plane of the nozzle. For a pressure ratio less than b but greater than e, two types of oblique shock wave patterns are observed, one with a central normal shock wave, as sketched for pr/p0 c, and one with only oblique waves, as sketched for pr/p0 d. The pressure ratios that result in oblique shock waves will not be considered here. As we move from d to e, the oblique shock waves become weaker and weaker until the isentropic flow is again realized at pr/p0 e with all shock waves absent. For pressure ratios below e a very complicated flow exists. The flow turns the corner at the nozzle exit rather abruptly due to expansion waves (isentropic waves to be considered in Section 9.8), then turns back due to the same expansion waves, resulting in a billowing out of the exhaust flow, as is visible from high-altitude satellite rocket engines. Let us work some examples for the converging–diverging nozzle; no new equations are necessary.

KEY CONCEPT The location of the shock wave is prescribed in student problems.


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pr a

Ve

p0

pe b

c p r /p 0 p/p 0

C d a

e D

b c d e f

f

x

(a) (b)

Fig. 9.12 Converging–diverging nozzle.

Example 9.7 A converging–diverging nozzle has a throat diameter of 5 cm and an exit diameter of 10 cm. The reservoir is the laboratory, maintained at atmospheric conditions of 20°C and 90 kPa absolute. Air is constantly pumped from a receiver so that a normal shock wave stands across the exit plane of the nozzle. Determine the receiver pressure and the mass flux. Solution Isentropic flow occurs from the reservoir, to the throat, to the exit plane in front of the normal shock wave at state 1. Supersonic flow occurs downstream of the throat making the throat the critical area. Hence 102 A1 4 52 A* Interpolation in the isentropic flow table (Table D.1) gives M1 2.94

p1 0.0298 p0

Hence the pressure in front of the normal shock is p1 p0 0.0298 90 0.0298 2.68 kPa


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From the normal shock table (Table D.2), using M1 2.94, we find that p2 9.918 p1 ⬖ p2 9.918 2.68 26.6 kPa This is the receiver pressure needed to orient the shock across the exit plane as shown for pr/p0 b in Fig. 9.12. To find the mass flux through the nozzle, we need only consider the throat. Recognizing that Mt 1, so that Vt ct, we can write pt kRTt pt At m˙ rt AtVt At 兹苶 RTt

k 冪莦 RT t

The isentropic flow table yields pt 0.5283 p0

Tt 0.8333 T0

Thus the mass flux becomes p 0.052 m˙ (0.5283 90 000) 4

冪莦莦 287 (0.8333 293) 1.4

0.417 kg앛s Remember, the pressure must be measured in pascals in the equation above. Check the units to make sure that the units on m˙ are kg/s.

Example 9.8 Air flows from a reservoir at 20°C and 200 kPa absolute through a 5-cm-diameter throat and exits from a 10-cm-diameter nozzle. Calculate the exit pressure pe needed to locate a normal shock wave at a position where the diameter is 7.5 cm. dt = 5 cm

d = 7.5 cm

p0

1

2

Ve

Shock wave

Fig. E9.8 Solution We will use the gas tables for this flow represented by the curve established by pr/p0 a in Fig. 9.12. The throat is a critical area since for a supersonic flow Mt 1. The area ratio is 7.5 2 A1 2.25 52 A*

(continued)


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For this area ratio we find, from Table D.1, that M1 2.33 Then from Table D.2, for this Mach number, we obtain p02 0.570 p01

M2 0.531 The reservoir pressure p0 p01. Thus

p02 0.570 200 114 kPa Isentropic flow occurs from state 2 immediately after the normal shock wave to the exit. Hence, for M2 0.531, we find from Table D.1 that A2 1.285 A* If Ae is the exit area, then Ae A2 Ae 102 1.285 2 2.284 A* A* A2 7.5 The Mach number and pressure ratio corresponding to this area ratio are pe 0.952 p0e

Me 0.265

For isentropic flow between the shock and the exit, we know that p02 p0e; thus pe p02 0.952 114 0.952 109 kPa Note the usefulness of the critical area ratio in obtaining the desired results.

Example 9.9 A pitot probe, the device used to measure the stagnation pressure in a flow, is inserted into an airstream and measures 300 kPa absolute, as shown in Fig. E9.9. The pressure in the flow is measured to be 75 kPa absolute. If the temperature at the stagnation point of the probe is measured as 150°C, determine the free-stream velocity V.

75 kPa

300 kPa

3

1 V

Pitot probe 2

Shock wave

Fig. E9.9


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Solution When a blunt object is placed in a supersonic flow a detached shock wave forms around the object, as it does around the front of the pitot probe shown. The flow that meets the front of the pitot probe at the stagnation point passes through a normal shock wave from state 1 to state 2; the subsonic flow at state 2 then decelerates isentropically to state 3, the stagnation point. For the isentropic flow from state 2 to state 3 we can use Eq. 9.3.13,

p3 k1 1 M 22 p2 2

冢

k앛(k 1)

冣

Across the normal shock we know that (see Eq. 9.4.12)

p2 2k k1 M 21 p1 k 1 k1

Also, the Mach numbers are related by Eq. 9.4.11, (k 1)M 21 2 M 22 2kM 21 k 1 The three equations above can be combined, with some algebraic manipulation, to yield the Rayleigh-pitot-tube formula for supersonic flows, namely,

k1 M 冣冢 2 2 1

k앛(k 1)

p3 p1 2kM 21 k 1 1앛(k 1) k1 k1

冢

冣

Substituting k 1.4, p3 300 kPa, and p1 75 kPa, we have (1.2M 12)3.5 300 (1.167M 12 0.1667)2.5 75

This can be solved by trial and error to give M1 1.65 (continued)


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The Mach number, after the normal shock is interpolated from the shock table to be M2 0.654 Using the isentropic flow table with this Mach number, we interpolate the temperature at state 2 to be T2 T3 0.921 423 0.921 389.6 K The temperature in front of the normal shock is found by using the normal shock table as follows: T2 1.423 T1 T2 ⬖ T1 274 K 1.423 Finally, the velocity before the normal shock is given by V1 M1c1 M1 兹苶 kRT1 1.65 兹苶 1.4 苶 287 苶 274 547 m앛s

9.6 VAPOR FLOW THROUGH A NOZZLE The flow of vapor through a nozzle forms a very important engineering problem. High-pressure steam flows through the nozzles of turbines in electrical generating plants; in this section we present the technique for analyzing such a problem. We recall, however, that vapor that is not substantially superheated does not behave very well as an ideal gas; the vapor tables must be consulted since cp and c√ cannot be assumed to be constant. Consider the problem of a superheated vapor entering the nozzle of Fig. 9.12. The flow would be isentropic unless a shock wave were encountered, and the expansion would be as sketched on the T–s diagram of Fig. 9.13. Assume that the flow initiates in a reservoir with stagnation conditions, flows through a throat indicated by state t, and exits out a diverging section to the exit at state e. Note that the exit state could very likely be in the quality region with the possibility of condensation of liquid droplets; however, for sufficiently high flow velocities there may not be sufficient time for the formation of the droplets and the associated heat transfer process. This produces a situation called supersaturation and a condition of metastable equilibrium exists; that is, the nonequilibrium state e is reached rather than the equilibrium state e . To estimate the temperature of the metastable state e, we assume that the steam behaves as an ideal gas,


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Sec. 9.6 / Vapor Flow Through a Nozzle T

455

Inset 0 p0 pe

pe pe

e′

e e

Fig. 9.13 vapor.

Condensation shock

s

Isentropic expansion of a

so that Tpk/(k 1) const. If the exit state is sufficiently far into the quality region, a condensation shock will be experienced, a phenomenon not considered in this book. Because of supersaturation it is possible to model the isentropic flow of a vapor through a nozzle with acceptable accuracy by considering the ratio of specific heats to be constant. For steam k 1.3 gives acceptable results over a considerable range of temperatures. The critical pressure ratio given by Eq. 9.3.14 for steam becomes

冢

p* 2 p0 k1

k앛(k 1)

冣

0.546

(9.6.1)

If saturated steam enters the nozzle, some liquid droplets will be formed and entrained by the vapor; provided that a condensation shock does not exist, a good approximation to the critical pressure ratio is 0.577 corresponding to k 1.14.

Example 9.10 Steam is to be expanded isentropically from reservoir conditions of 300°C and 800 kPa absolute to an exit condition of 100 kPa absolute. If supersonic flow is desired, calculate the necessary throat and exit diameters if a mass flux of 2 kg/s is demanded. Solution From the steam tables (found in any thermodynamics textbook) we find that s0 se 7.2336 kJ앛kg K

h0 3056.4 kJ앛kg

To estimate the temperature of the metastable exit state, we use pe Te T0 p0

(k 1)앛k

冢冣

冢冣

100 593 800

0.3앛1.3

367 K

or

94°C (continued)


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Using the steam tables at this temperature, we interpolate, using the exit quality xe, to find that 7.2336 1.239 6.90xe ⬖ xe 0.968 Thus we have the enthalpy and specific volume at the exit: he 394 0.968 2273 2594 kJ앛kg √e 0.001 0.968 (2.06 0.001) 1.99 m3앛kg Using the energy equation, the exit velocity is estimated as follows: QQQ QQQ QQO

0

V 02 V e2 h0 he 2 2 ⬖ Ve 兹苶 2(h0 苶 he) 兹2(3056 苶苶 259苶 4) 1000 苶 961 m앛s

where the 1000 converts kJ to J. From the definition of mass flux we have me re AeVe p d 2e 1 2 961 4 1.99 ⬖ de 0.0726 m

or

7.26 cm

To determine the diameter of the throat, we recognize the throat to be the critical area; thus Eq. 9.6.1 gives p* 0.546 p0 437 kPa Using this pressure and s* s0 7.2336 kJ/kg K we could use the steam tables to find h* and √*; the energy equation would then allow us to find V* and thus dt. However, a simpler, approximate technique, assuming constant specific heats, is to use Eq. 9.3.18 with k 1.3 and obtain the following: m˙ p0 A*

1.3 2.3 冪莦 RT 冢 2 冣

2.3앛0.6

0

pd 2t 2 800 000 4

0.585 冪莦 462 莦 573 1.3

⬖ dt 0.049 m or 4.9 cm This is reasonable since we have already assumed constant specific heats in predicting Te. Obviously, the above is approximate; using k 1.3 does, however, give reasonable predictions.

9.7 OBLIQUE SHOCK WAVE In this section we investigate the oblique shock wave, a finite-amplitude wave that is not normal to the incoming flow. The flow approaching the oblique shock wave will be assumed to be in the x-direction.After the oblique shock the velocity vector


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Sec. 9.7 / Oblique Shock Wave 457

Oblique shock V2

V1

V2

Oblique shock

V1

V2 (a)

(b)

Fig. 9.14 Oblique shock waves in a supersonic flow: (a) flow over a symmetrical wedge; (b) flow in a corner.

will have a component normal to the flow direction. We will continue to assume that the flow before and after the oblique shock is uniform and steady. Oblique shock waves form on the leading edge of a supersonic airfoil or in an abrupt corner, as sketched in Fig. 9.14. Oblique shock waves may also be found on axisymmetric bodies such as a nose cone or a bullet traveling at supersonic speeds. In this book, we consider only plane flows. The function of the oblique shock wave is to turns the flow so that the velocity vector V2 is parallel to the plane wall. The angle between the two velocity vectors introduces another variable into our analysis. The problem remains solvable, however, with the additional tangential momentum equation. To analyze the oblique shock wave, consider a control volume enclosing a portion of the shock as shown in Fig. 9.15. The velocity vector upstream is assumed to be in the x-direction only; the oblique shock wave makes an angle b with the upstream velocity vector and turns the flow through the deflection angle or wedge angle u so that V2 is parallel to the wall. The components of the velocity vectors are shown normal and tangential to the oblique shock wave. The tangential components do not cause fluid to flow through the shock; hence the continuity equation, with A1 A2, gives r1V1n r2V2n

KEY CONCEPT The oblique shock wave turns the flow so that the velocity vector is parallel to the plane wall.

(9.7.1)

where the normal components V1n and V2n are displayed in Fig. 9.15. The pressure forces act normal to the oblique shock and produce no tangential components. Thus the momentum equation expressed in the tangential direction requires that the tangential momentum into the control volume equal the tangential momentum leaving the control volume; that is, m˙ 1V1t m˙ 2V2t

(9.7.2)

˙1 m ˙ 2, we demand that or, using m V1t V2t

(9.7.3)


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Control volume

β

β–θ

V1t V1n

V2 V2n

V1

β

Oblique shock wave

V2t

θ x

Fig. 9.15

Control volume enclosing a small portion of an oblique shock wave.

The normal momentum equation takes the form 2 p1 p2 r 2V 22n r1V 1n

(9.7.4)

The energy equation, using V 2 V n2 V 2t , may be written as 2 V 21n k p1 V 2n k p2 2 k 1 r1 2 k 1 r2

KEY CONCEPT The tangential components of the two velocity vectors do not enter the equations.

(9.7.5)

where the tangential component terms have been canceled from both sides. Note that the tangential components of the two velocity vectors do not enter the continuity, normal momentum, or energy equations, the three equations used in the solution of the normal shock wave. Hence we may substitute V1n and V2n for V1 and V2, respectively, in the normal shock wave equations and obtain a solution. Either the normal shock wave equations or the normal shock wave table (Table D.2) may be used. We also replace M1 and M2 with M1n and M2n, respectively. It is useful to relate the oblique shock angle b to the deflection angle u. Using the continuity equation (9.7.1), with reference to Fig. 9.15, there results r2 V1n V1t tan b tan b r1 V2n V2t tan(b u) tan(b u)

(9.7.6)

From the normal shock wave equations (9.4.12) and (9.4.13) we can find the density ratio to be 2 r2 p2T1 (k 1)M 1n 2 r1 p1T2 (k 1)M 1n 2

(9.7.7)


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Sec. 9.7 / Oblique Shock Wave 459

Substituting this into Eq. 9.7.6 gives tan b 2 tan(b u) k 1 k1 M 12 sin2 b

冤

冥

(9.7.8)

For a flow with a given M1, this equation relates the oblique shock angle b to the wedge or corner angle u. The three variables b, u, and M1 in the equation above are often plotted as in Fig. 9.16. We can observe several phenomena by studying the figure. • For a specified upstream Mach number M1 and a given wedge angle u there are two possible oblique shock angles b, the larger one corresponding to a “strong” shock and the smaller corresponding to a “weak” shock. • For a given wedge angle u there is a minimum Mach number for which there is only one oblique shock angle b. • For a given wedge angle u, if M1 is less than the minimum for that particular curve, no oblique shock wave exists and the shock wave becomes detached, as shown in Fig. 9.17. Also, for a given M1 there is a sufficiently large u that will result in a detached shock wave. The pressure rise across the oblique shock wave determines whether a weak shock or a strong shock occurs. For a relatively small pressure rise a weak shock will occur with M2 1. If the pressure rise is relatively large a strong shock occurs with M2 1. Note that for the detached shocks around bodies, a normal shock exists for the stagnation streamline; this is followed away from the stagnation point by a strong oblique shock, then a weak oblique shock, and eventually a Mach wave. For blunt bodies moving at supersonic speeds the shock wave is always detached.

80 M2 < 1

Strong shock

70

M2 = 1 60

θ = 30°

Weak shock

M2 > 1

β 50

θ = 35° θ = 25°

40

θ = 20° θ = 15°

30

θ = 10° θ = 5°

20

θ = 0° 1.5

2.0

2.5

3.0

3.5

M1

Fig. 9.16

Oblique shock wave relationships for k 1.4.


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Weak shock Strong shock

Weak shock Strong shock

θ

M2 > 1

M2 < 1 M2 < 1

M2 > 1

(a)

(b)

Fig. 9.17 Detached shock waves: (a) flow around a wedge; (b) flow around a blunt object.

Example 9.11 Air flows over a wedge with M1 3, as shown in Fig. E9.11. A weak shock reflects from the wall. Determine the values of M3 and b 3 for the reflected wave.

20° M1 = 3.0

β1

θ1

V2

θ2

V1

β2

V3

β3

Fig. E9.11

Solution From Fig. 9.16 with u1 10° and M1 3.0, we find for the weak shock that b1 27.5°. This yields M1n 3 sin 27.5° 1.39 From the shock table we interpolate to find M2n 0.744 M2 sin(27.5° 10°) ⬖ M2 2.48


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Sec. 9.8 / Isentropic Expansion Waves 461

The reflected shock must again turn the flow through an angle of 10°, that is, u2 10°. For this wedge angle and M2 2.48 from Fig. 9.16 for a weak shock, we see that b2 33°. This results in M2n 2.48 sin 33° 1.35 From the shock table M3n 0.762 M3 sin 23° ⬖ M3 1.95 The desired angle is calculated to be b3 b2 10° 23° Note that Fig. 9.16 does not allow precise calculations. Equation 9.7.8 could be used, by trial and error, to improve the accuracy of the b’s and hence the quantities that follow.

9.8 ISENTROPIC EXPANSION WAVES In this section we consider the supersonic flow around a convex corner, as shown in Fig. 9.18. Let us first attempt to accomplish such a flow with the finite amplitude wave of Fig. 9.18a. The flow must turn the angle u so that V2 is parallel to the wall. The tangential component must be conserved because of momentum conservation. This would result in V2 V1, as is obvious from the sketch. This would be the situation if a subsonic flow, M1n 1, could experience a finite increase to a supersonic flow, M2n 1. This, of course, is impossible because of the second law, as was noted in the discussion associated with Fig. 9.10. Consequently, we consider the turning of a flow around a convex corner using a finite wave as an impossibility. Consider a second possible mechanism that would allow the flow to turn the corner, a fan composed of an infinite number of Mach waves, emanating from the

Expansion fan

Finite wave V1

V1t

V2t

V1

θ

(a) Single finite wave

M1

θ

V2

V2 M2

(b) Infinite number of Mach waves

Fig. 9.18 Supersonic flow around a convex corner.


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Control volume Mach wave Vt

Vn

μ + dθ

V

μ

1 sin μ = –– M

Vn + dVn

M2− 1 cosμ = ––––––– M

dθ

V + dV Vt

Fig. 9.19 A single Mach wave.

corner, as shown in Fig. 9.18b. The second law would not be violated with such a mechanism since each Mach wave is an isentropic wave. We will determine the effect of a single Mach wave on the flow and then integrate to obtain the total effect. Figure 9.19 shows the infinitesimal velocity change due to a single Mach wave. For the control volume enclosing the Mach wave, we know that the tangential momentum is conserved; thus the tangential velocity component remains unchanged as shown, as in the oblique shock wave. From the triangles in the figure we can write Vt V cos m (V dV) cos(m du)

(9.8.1)

Since du is small, this becomes,1 using cos(m du) cos m du sin m, V sin m du cos m dV

(9.8.2)

Substituting sin m 1/M (see Eq. 9.2.13) and cos m (兹苶 M2 1苶)/M, we have dV du 兹苶 M 2 苶1 V

(9.8.3)

The relationship V M 兹kRT 苶 can be differentiated and rearranged to give dV d M 1 dT V M 2 T The energy equation, in the form

(9.8.4)

1 2 V kRT/(k 1) const., may also be dif2

ferentiated to yield dV 1 dT 2 0 V (k 1)M T

(9.8.5)

Recall the trigonometric identity, cos(a b) cos a cos b sin a sin b. Then using cos du 1 and sin du du, we have cos(m du) cos m du sin m.

1


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Sec. 9.8 / Isentropic Expansion Waves 463

Eliminating dT/T by combining the two preceding equations, there results

dV 2 dM 2 V 2 (k 1)M M

(9.8.6)

This can be substituted into Eq. 9.8.3, allowing a relationship between u and M to be obtained. We have

2 苶 2兹M 苶1 dM du 2 2 (k 1)M M

(9.8.7)

This can be integrated, using u 0 at M 1, to provide a relationship between the resulting Mach number (M2 in Fig. 9.18) and the angle, provided that the incoming Mach number is unity; the relationship is

k1 u k1

冢

冣

1앛2

k1 tan1 (M2 1) k1

冤

冥

1앛2

tan1(M2 1)1앛2

(9.8.8)

The angle u, which is a function of M, is the Prandtl–Meyer function. It is tabulated for k 1.4 in Table D.3 so that trial-and-error solutions to Eq. 9.8.8 for M are not necessary. Other changes that may be desired, such as the pressure or temperature changes, can be found from the isentropic-flow equations. The collection of Mach waves that turn the flow are referred to as an expansion fan. We will see, in working the examples and problems, that both the Mach number and the velocity increase as supersonic flow turns the convex corner. The flow remains attached to the wall as it turns the corner, even for large angles, a phenomenon not observed in a subsonic flow; a subsonic flow would separate from the abrupt corner, even for small angles. If we substitute M in Eq. 9.8.8, we find the angle of maximum turning to be u 130.5°. This would mean that both the temperature and pressure would be absolute zero; obviously, the gas would become a liquid before this would be possible. The angle of 130.5° is, however, an upper limit. The point is, rather large turning angles are possible in supersonic flows, angles that may exceed 90°, a rather surprising result. This introduces a design constraint on the exhaust nozzle of rocket engines that exhaust into the vacuum of space; the exhaust gases may turn through an angle so that impingement on the spacecraft body would occur if not designed properly.

Prandtl–Meyer function: The angle u through which the supersonic flow turns.

KEY CONCEPT The flow remains attached to the wall as it turns the corner, even for large angles.

KEY CONCEPT Rather large turning angles are possible in supersonic flows.


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Example 9.12 Air at a Mach number of 2.0 and a temperature and pressure of 500°C and 200 kPa absolute, respectively, flows around a corner with a convex angle of 20° (Fig. E9.12). Find M2, p2, T2, and V2.

M1 = 2.0 M=1

μ 1 = 30°

90°

μ1

μ2

26.4° = θ1

M2

20°

Fig. E9.12

Solution Table D.3 uses M 1 as a reference condition; thus we visualize the flow as originating from a flow with M 1 and turning through the angle u1 to M1 2, as shown in the sketch. From the table, adding an additional 20° to the deflection angle, we find that u2 46.4°. This would be equivalent to the flow at M 1 turning a convex corner with u 46.4°. Since the flow is isentropic, we can simply superimpose in this manner. Now, for an angle of u 46.4° from the table, we find that M2 2.83 From the isentropic flow table (Table D.1) we find that p0 p2 p2 p1 p1 p0 1 200 0.0352 55.1 kPa 0.1278 T0 T2 T2 T1 T1 T0 1 773 0.3844 534.8 K 0.5556

or

261.8°C

The velocity V2 is found to be V2 M2 兹kRT 苶2 2.83 兹1.4 苶287 苶苶 534.8 1312 m앛s


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9.9 SUMMARY The zone of silence of a supersonic object that produces only Mach waves exists outside a cone with included angle a found from 1 sin a M

(9.9.1)

where the Mach number is M V/c and c 兹kRT 苶. The relationship dV dA (M2 1) V A

(9.9.2)

allows us to predict how subsonic and supersonic flows behave in converging and diverging nozzles. The mass flux through a nozzle with throat area A* where M* 1 is given by ˙ p0 A* m

k k1 冪莦 RT 冢 2 冣

(k 1)앛2(1 k)

(9.9.3)

0

where p0 and T0 are the reservoir conditions. Temperature, pressure, and velocity in an isentropic flow are found from T0 k1 1 M2, T 2

p0 k1 1 M2 p 2

冢

k앛(k 1)

冣

,

V2 cp(T0 T) (9.9.4) 2

The flow variables across a normal shock are found from continuity, momentum, and energy: r1V1 r 2V2,

p1 p2 r1V1(V2 V1),

p2 p1 k 0 2 k 1 r2 r1 V 22

V 21

冢

冣

(9.9.5)

Rather than solve the above equations, we often use the normal-shock flow Table D.2. Across an oblique shock the tangential velocity component does not change. The normal velocity component Vn simply replaces the velocity V in the normalshock wave equations above, and Table D.2 may be used with Mn replacing M. The wedge angle u through which the flow is turned is related to the oblique shock angle b by

冤

冥

tan b 2 tan(b u) k 1 k1 M21 sin2b

(9.9.6)


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or Fig. 9.16 may be used to avoid a trial-and-error solution if b is desired. In an expansion wave, the angle u through which a flow with M 1 can be turned is the Prandtl–Meyer function: k1 u k1

冢

冣

1앛2

k1 tan1 (M2 1) k1

冤

冥

1앛2

tan1(M2 1)1앛2

(9.9.7)

If we know u and desire M, a trial-and-error solution is required. This is avoided by using Table D.3.

PROBLEMS 9.1

In thermodynamics cp for air is often used as 0.24 Btu/lbm-°R. From Table B.4 cp is 6012 ft-lb/slug-°R. Show these two values to be equal. Also, calculate c√ in both sets of units. (Note: The value of cp used in thermodynamics and in fluid mechanics is the same when using the SI system of units.)

9.2 9.3 9.4

Show that cp Rk/(k 1). Show that Eqs. 9.1.10 follow from Eqs. 9.1.9 and 9.1.7. Verify the various forms of the energy equation in Eqs. 9.1.3, 9.1.4, and 9.1.5.

Speed of Sound 9.5 9.6

Show that Eq. 9.2.8 follows from Eq. 9.2.7. Show that the velocity of sound in a high-frequency wave (i.e., a dog whistle) travels with speed c 兹RT 苶 if the process is assumed to be isothermal. 9.7 Show that for a small adiabatic disturbance in a steady flow, the energy equation takes the form h c V. 9.8 Verify that the speed of propagation of a small wave through water is about 1450 m/s. 9.9 Two rocks are slammed together on one side of a lake. An observer on the other side with his head under water “hears” the disturbance 0.6 s later. How far is it across the lake? 9.10 You and a friend are standing 10 m apart in waistdeep water. You slam two rocks together under water. How long after the rocks are slammed together does your friend hear the interaction if your friend’s head is underwater? 9.11 Calculate the Mach number for an aircraft if it is flying at: (a) Sea level at 200 m/s (b) 15,000 ft at 600 fps (c) 10 000 m at 200 m/s (d) 60,000 ft at 600 fps (e) 35 000 m at 200 m/s

9.12

A wood chopper is chopping wood some distance away. You carefully observe, using your numerical wrist timer, that it takes 1.21 s for the sound of the axe to reach your ears. Calculate the distance you are from the chopper if the temperature is 10°C.

9.13

You see a bolt of lightning and 2 s later you hear the sound of thunder. How far away did the lightning bolt strike? Use: (a) SI units (b) English units

9.14

A needle-nosed projectile passes over you on a military testing field at a speed of 1000 m/s (Fig. P9.14). You know that it has an elevation of 1000 m where T 10°C. How long after it passes overhead will you hear its sound? How far away will it be? Calculate its Mach number. 1000 m/s

1000 m

L

Fig. P9.14


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Problems 467 9.15

A special camera is capable of showing the Mach angle of a sharp-nosed bullet passing through the test section of a wind tunnel. If the Mach angle is measured to be 22°, calculate the speed of the bullet in: (a) m/s (b) ft/sec Assume standard conditions.

9.16

A small-amplitude wave passes through the standard atmosphere at sea level with a pressure rise of 0.3 psf. Estimate the associated induced velocity and temperature rise.

Isentropic Flow 9.17

Provide all the steps needed to proceed from: (a) Eq. 9.3.2 to Eq. 9.3.3 (b) Eq. 9.3.5 to Eq. 9.3.10 (c) Eq. 9.3.11 to Eq. 9.3.12 (d) Eq. 9.3.16 to Eq. 9.3.18 (e) Eq. 9.3.17 to Eq. 9.3.19 p0

9.18

A pitot probe, an instrument that measures stagnation pressure, is used to determine the velocity of an airplane. Such a probe, attached to an aircraft, measures 10 kPa. Determine the speed of the aircraft if it is flying at an altitude of: (a) 3000 m (b) 10 000 m Assume an isentropic process from the free stream to the stagnation point.

9.19

A pitot probe, used to measure the stagnation pressure, indicates a pressure of 4 kPa at the nose of a surface vehicle traveling in 15°C–atmospheric air. Calculate its speed assuming: (a) Isentropic flow (b) Incompressible flow Compute the percent error for part (b).

9.20

9.21

A converging nozzle with an exit diameter of 2 cm is attached to a reservoir maintained at 25°C and 200 kPa absolute. Using equations only, determine the mass flux of air if the receiver pressure is: (a) 100 kPa absolute (b) 130 kPa absolute The converging nozzle of Fig. P9.21 is attached to a reservoir maintained at 70°F and 30 psia. Using equations only, determine the mass flux of air if the receiver pressure is: (a) 15 psia (b) 20 psia

T0

Fig. P9.21 9.22 Rework Problem 9.20 using the isentropic flow table. 9.23 Rework Problem 9.21 using the isentropic flow table. 9.24

Air flows from a reservoir (T0 30°C, p0 400 kPa absolute) out a converging nozzle with a 10-cm-diameter exit. What exit pressure would just result in Me 1? Determine the mass flux for this condition. Use the isentropic flow table.

9.25 Air flows from a converging nozzle attached to a reservoir with T0 10°C. What reservoir pressure is necessary to just cause Me 1 if the 6-cm-diameter nozzle exits to atmospheric pressure? Calculate the mass flux for this condition. 9.26

Air flows from a converging nozzle attached to a reservoir with T0 40°F. If the 2.5-in.-diameter nozzle exits to atmospheric pressure, what reservoir pressure is necessary to just cause Me 1? Calculate the mass flux for this condition. Now double the reservoir pressure and determine the increased mass flux.

9.27

A 25-cm-diameter air line is pressurized to 500 kPa absolute and suddenly bursts (Fig. P9.27). The exit


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Chapter 9 / Compressible Flow area is later measured to be 30 cm2. If 6 min elapsed before the 10°C air was turned off, estimate the cubic meters of air that was lost.

25 cm dia. 10 °C

Air

Fig. P9.27 9.28

A converging nozzle is attached to a reservoir containing helium with T0 27°C and p0 200 kPa absolute. Determine the receiver pressure that will just give Me 1. Now attach a diverging section with a 15-cm-diameter exit to the 6-cm-diameter throat. What is the highest receiver pressure that will give Mt 1? 9.29 A venturi tube is used to measure the mass flux of air in a pipe by reducing the diameter from 10 cm to 5 cm and then back to 10 cm. The pressure at the inlet is 300 kPa and at the minimum diameter section it is 240 kPa. If the upstream temperature is 20°C, determine the mass flux.

9.30

Air enters a converting-diverging nozzle from a reservoir at 200 kPa (gage) and 22°C. the area at the throat is 9.7 cm2, and at the exit it is 13 cm2. Considering isentropic flow throughout with M 1 at the throat, determine the velocity of air at the nozzle exit. 9.31 Consider the nozzle flow of Problem 9.30, but in this case the Mach number at the throat is 0.72. Determine the velocity at the nozzle exit for the given condition. 9.32 The mass flux of air flowing through the pipe of Fig. P9.32 is desired. The pipe is reduced in diameter from 4 in. to 2 in. and then back to 4 in. The pressure at the inlet is 45 psi and at the minimum diameter section it is 36 psi. If the upstream temperature is 60°F, determine the mass flux.

45 psi

36 psi

2 diameter 4 diameter

Fig. P9.32 9.37 Air flows from a reservoir maintained at 30°C and 200 kPa absolute through a converging–diverging nozzle that has a 10-cm-diameter throat. Determine the diameter where M 3. Use equations only. 9.34 Air flows in a converging–diverging nozzle from a 9.38 reservoir maintained at 20°C and 500 kPa absolute. The throat and exit diameters are 5 cm and 15 cm, respectively. What two receiver pressures will result in M 1 at the throat if isentropic flow occurs throughout? Use equations only. 9.35 Rework Problem 9.34 using the isentropic flow table. Reservoir 3.5 MPa 9.36 Air flows from a nozzle with a mass flux of 320 K 1.0 slug/sec. If T0 607°F, p0 120 psia, and pe 15 psia. Calculate the throat and exit diameters for an isentropic flow. Also, determine the exit velocity.

9.33

Air flows from a reservoir, maintained at 20°C and 2 MPa absolute, and exits from a nozzle with Me 4. The receiver pressure is then raised until the flow is just subsonic throughout the entire nozzle. Calculate this receiver pressure. A small wind tunnel uses compressed air from a reservoir at 3.5 MPa and 320 K. The air flows isentropically from the reservoir through a nozzle, as shown in Fig. P9.38. If the Mach number in the test section is 2.8, calculate the pressure and velocity in the test section Test section M = 2.8

Fig. P9.38


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Problems 469 9.39 Consider isentropic air flow from a reservoir (where p0 600 kPa, T0 30°C) through a convergingdiverging nozzle. At a section in the converging part of the nozzle before the throat, the Mach number is 0.50, and the cross-sectional area is 12.4 cm2. (a) If the area at the throat is 10 cm2, calculate the pressure, temperature, and velocity at the throat. (b) If M 1 at the throat, what should be the receiver pressure to just produce supersonic flow at the exit? (c) If the Mach number is 2.0 at the exit, what should be the exit area and the mass flux? 9.40 Air at 30°C flows in a 10-cm-diameter pipe at a velocity of 150 m/s. A venturi tube is used to measure the flow rate. What should be the minimum diameter of the tube so that supersonic flow does not occur? 9.41 For a nozzle efficiency of 96%, rework Problem 9.23. 9.42 Nitrogen enters a diffuser at 100 kPa absolute and 100°C with a Mach number of 3.0. The mass flux is 10 kg/s and the exit velocity is small. Sketch the diffuser, then determine the throat area and the exit pressure and temperature, assuming isentropic flow. 9.43 Nitrogen enters a diffuser at 15 psia absolute and 200°F with a Mach number of 3.0. The mass flux is 0.2 slug/sec and the exit velocity is small. Sketch the diffuser, then determine the throat area and the exit pressure and temperature, assuming isentropic flow. 9.44 A rocket has a mass of 80 000 kg and is to be lifted vertically from a platform with six nozzles exiting

exhaust gases with Te 1000°C. What should the exit velocity be from each 50-cm-diameter nozzle if the exhaust gases are assumed to be carbon dioxide? 9.45 A 100-kg man straps a small air-breathing jet engine on his back and just lifts off the ground vertically (Fig. P9.45). The engine has an exit area of 200 cm2. With what velocity must the 600°C exhaust gases exit the engine?

Fig. P9.45 9.46

9.47

A converging–diverging nozzle is bolted into a reservoir at a diameter of 40 cm. The throat and exit diameters are 5 cm and 10 cm, respectively. If T0 27°C and pe 100 kPa absolute and isentropic flow of air occurs throughout the supersonic nozzle, calculate the force necessary to hold the nozzle onto the reservoir. What is the maximum speed in (a) m/s and (b) mph that an aircraft can have during take-off or landing if the airflow around the aircraft is to be modeled as an incompressible flow? Allow a 3% error in the pressure from free stream to the stagnation point. Assume standard conditions.

Normal Shock 9.48

The pressure, temperature, and velocity before a normal shock wave are 80 kPa absolute, 10°C, and 1000 m/s, respectively. Calculate M1, M2, p2, T2, and r2 for air. Use: (a) Basic equations (b) The normal shock table

9.49

The pressure, temperature, and velocity before a normal shock wave are 12 psia, 40°F, and 3000 ft/sec, respectively. Calculate M1, M2, p2, T2, and r2 for air. Use: (a) Basic equations (b) The normal shock table

9.50

Derive the Rankine–Hugoniot relationship, (k 1) p2앛p1 k 1 r2 (k 1) p2앛p1 k 1 r1

which relates the density ratio to the pressure ratio across a normal shock wave. Find the limiting density ratio for air across a strong shock for which p2/p1 1. 9.51 An explosion occurs just above the surface of the Earth, producing a shock wave that travels radially outward. At a given location it has a Mach number of 2.0. Determine the pressure just behind the shock and the induced velocity. 9.52 Air at 200 kPa absolute and 20°C passes through a normal shock wave with strength so that M2 0.5. Calculate V1, p2, and r2. 9.53 Air at 30 psia and 60°F passes through a normal shock wave with strength so that M2 0.5. Calculate V1, p2, and r2. 9.54 A blunt object travels at 1000 m/s at an elevation of 10 000 m. The flow approaching the stagnation


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Chapter 9 / Compressible Flow point passes through a normal shock wave and then decelerates isentropically to the stagnation point. Calculate p0 and T0 at the stagnation point.

9.55 A pitot probe is inserted in an airflow in a pipe in which p 800 kPa absolute, T 40°C, and M 3.0 (Fig. P9.55). What pressure does it measure?

p

800 kPa

M = 3.0 Shock

Fig. P9.55 9.56

Air flows from a 25°C reservoir to the atmosphere through a nozzle with a 5-cm-diameter throat and a 10-cm-diameter exit. What reservoir pressure will just result in M 1 at the throat? Also, calculate the mass flux. Maintaining this reservoir pressure, reduce the throat diameter to 4 cm and determine the resulting mass flux. Sketch the pressure distribution as in Fig. 9.12.

9.57

Air flows from a 20°C reservoir to the atmosphere through a nozzle with a 5-cm-diameter throat and a 10-cm-diameter exit. What reservoir pressure is necessary to locate a normal shock wave at the exit? Also, calculate the velocity and pressure at the throat, before the shock, and after the shock.

9.58

Air flows from a 60°F reservoir to the atmosphere through a nozzle with a 2-in.-diameter throat and a 4-in.-diameter exit. What reservoir pressure is necessary to locate a normal shock wave at the exit? Also, calculate the velocity and pressure at the throat, before the shock, and after the shock.

9.59

Air flows from a reservoir maintained at 25°C and 500 kPa absolute. It flows through a nozzle with throat and exit diameters of 5 cm and 10 cm, respectively. What receiver pressure is necessary to locate a normal shock wave at a location where the diameter is 8 cm? Also, determine the velocity before the shock and at the exit.

Vapor Flow 9.60

Steam flows at a rate of 4 kg/s from reservoir conditions of 400°C and 1.2 MPa absolute through a converging–diverging nozzle to the atmosphere. Determine the throat and exit diameters if supersonic, isentropic flow exists throughout the diverging section.

9.61 Steam flows from reservoir conditions of 350°C and 1000 kPa absolute to the atmosphere at the rate of 15 kg/s. Estimate the converging nozzle exit diameter.

9.62

Steam flows from reservoir conditions of 700°F and 150 psia to the atmosphere at the rate of 0.25 slug/sec. Estimate the converging nozzle exit diameter.

9.63 A header supplies steam at 400°C and 1.2 MPa absolute to a set of nozzles with throat diameters of 1.5 cm. The nozzles exhaust to a pressure of 120 kPa absolute. If the flow is approximately isentropic, calculate the mass flux and the exit temperature.

Oblique Shock Wave 9.64

An airflow with velocity, temperature, and pressure of 800 m/s, 30°C, and 40 kPa absolute, respectively, is turned with an oblique shock wave emanating from the wall, which makes an abrupt 20° corner.

(a) (b) (c)

Find the downstream Mach number, pressure, and velocity for a weak shock. Find the downstream Mach number, pressure, and velocity for a strong shock. If the concave corner angle were 35°, sketch the corner flow situation.


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Problems 471 9.65

Two oblique shocks intersect as shown in Fig. P9.65. Determine the angle of the reflected shocks if the airflow must leave parallel to its original direction. Also find M3.

M1 = 2

9.68

β 60°

A supersonic inlet can be designed to have a normal shock wave oriented at the inlet, or a wedge can be used to provide a weak oblique shock wave, as shown in Fig. P9.68. Compare the pressure p3 from the flow shown to the pressure that would exist behind a normal shock wave with no oblique shock.

M3 Oblique shock

Fig. P9.65 9.66

9.67

M1 = 3

An oblique shock wave at a 35° angle is reflected from a plane wall. The upstream Mach number M1 is 3.5 and T1 0°C. Find V3 after the reflected oblique shock wave for the airflow.

p1 = 40 kPa absolute

An oblique shock wave at a 35° angle is reflected from a plane wall. The upstream Mach number M1 is 3.5 and T1 30°F. Find V3 after the reflected oblique shock wave for the airflow.

M3

M2

20°

M3

Normal shock

Fig. P9.68

Expansion Waves 9.69

9.70

A supersonic airflow with M1 3, T1 20°C, and p1 20 kPa absolute turns a convex corner of 25°. Calculate M2, p2, T2, and V2 after the expansion fan. Also calculate the included angle of the fan. A supersonic airflow with M1 2, T1 0°C, and p1 20 kPa absolute turns a convex corner. If M2 4, what angle u should the corner have? Also calculate T2 and V2.

9.71

A supersonic airflow with M1 2, T1 30°F, and p1 5 psia turns a convex corner. If M2 4, what angle u should the corner have? Also calculate T2 and V2.

9.72

The flat plate shown in Fig. P9.72 is used as an airfoil at an angle of attack of 5°. Oblique shock waves and expansion fans allow the air to remain attached to the plate with the flow behind the airfoil parallel to the original direction. Calculate (a) The pressures on the upper and lower sides of the plate (b) The downstream Mach numbers M2u and M2l (c) The lift coefficient defined by CL lift앛(12 r1V 21A). Note that r1V 21 kM 21r1

M1 = 2.5


Mu

M2u

Ml

M2l

Fig. P9.72 9.73 The supersonic airfoil shown in Fig. P9.73 is to fly at zero angle of attack. Calculate the drag coefficient CD drag앛(21 r1V 12A). Note that r1V 12 kM 12 r1. M1 = 4 5°

5°


Fig. P9.73 9.74

The airfoil in Problem 9.73 flies at an angle of attack of 5°. Determine the lift and drag coefficients, CL and CD. See Problems 9.72 and 9.73 for definitions of CL and CD.


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The Wahluke Branch Canal, part of the Columbia Basin Project, is an example of an engineered canal. Water is delivered over a distance of many kilometers. (U.S. Bureau of Reclamation)


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10 Flow in Open Channels Outline 10.1 Introduction 10.2 Open-Channel Flows 10.2.1 Classification of Free-Surface Flows 10.2.2 Significance of Froude Number 10.2.3 Hydrostatic Pressure Distribution 10.3 Uniform Flow 10.3.1 Channel Geometry 10.3.2 Equation for Uniform Flow 10.3.3 Most Efficient Section 10.4 Energy Concepts 10.4.1 Specific Energy 10.4.2 Use of the Energy Equation in Transitions 10.4.3 Flow Measurement

10.5 Momentum Concepts 10.5.1 Momentum Equation 10.5.2 Hydraulic Jump 10.5.3 Numerical Solution of Momentum Equation 10.6 Nonuniform Gradually Varied Flow 10.6.1 Differential Equation for Gradually Varied Flow 10.6.2 Water Surface Profiles 10.6.3 Controls and Critical Flow 10.6.4 Profile Synthesis 10.7 Numerical Analysis of Water Surface Profiles 10.7.1 Standard Step Method 10.7.2 Irregular Channels 10.7.3 Computer Method for Regular Channels 10.7.4 Direct Integration Methods 10.8 Summary

Chapter Objectives The objectives of this chapter are to: ▲ Describe various types of open-channel, free-surface flows ▲ Apply the Chezy–Manning equation with various geometric cross-sections for uniform flow applications ▲ Derive energy and momentum principles for rapidly varied flow situations ▲ Derive the differential equation for nonuniform gradually varied flow ▲ Present the method of profile synthesis—the qualitative description of openchannel flow, including establishment of controls and classification of water surface profiles ▲ Develop and present methods to numerically compute gradually varied flow ▲ Present numerous examples of uniform, rapidly varied, and gradually varied flow ▲ Detail several examples of profile synthesis and numerical analysis of complex open-channel flow with emphasis on design application 473


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Chapter 10 / Flow in Open Channels

10.1 INTRODUCTION

Free surface: The interface between the air and the upper layer of water.

Free-surface flow is probably the most commonly occurring flow phenomenon that we encounter on the surface of the earth. Ocean waves, river currents, and overland flow of rainfall are examples that occur in nature. Man-made situations include flows in canals and culverts, drainage over impervious materials such as roofs and parking lots, and wave motion in harbors. In all of these situations, the flow is characterized by an interface between the air and the upper layer of water, which is termed the free surface. At the free surface, the pressure is constant, and for nearly all situations, it is atmospheric. In such a case, the hydraulic grade line and the liquid free surface coincide. In engineering practice, most open channels convey water as the fluid. However, the principles developed and implemented in this chapter are also applicable for other liquids that flow with a free surface. Generally, the elevation of the free surface does not remain constant; it can vary along with the fluid velocities. Another complexity is that the flow is often three-dimensional. Fortunately, there are many instances in which two-dimensional and even one-dimensional simplifications can be made. The flow patterns in estuaries1 and lakes under certain circumstances can be treated as two-dimensional in the horizontal plane, averaged vertically over the depth. River and channel flows are usually treated as one-dimensional with respect to the position coordinate along the streambed. In this chapter we restrict our consideration to one-dimensional flows. Figure 10.1a shows a representative centerline velocity distribution in a channel. The velocity profile is three-dimensional at a given cross section (Fig. 10.1b). The boundary shear stress is nonuniform; at the free surface the shear

ν

y

x (a) Velocity contours

V Secondary currents (b)

(c)

Fig. 10.1 Free-surface flow: (a) centerline velocity distribution; (b) cross section; (c) one-dimensional model. 1

An estuary is the lower course of a river that is influenced by the ocean tides.


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Sec. 10.2 / Open-Channel Flows 475

stress is negligible, yet the shear stress varies around the wetted perimeter. In some circumstances the presence of secondary currents will force the maximum velocity to occur slightly below the free surface. By convention, y is defined as the depth from the deepest location to the free surface; note that y is not a coordinate. The mean velocity is given by the relation 1 V A

冕

√ dA

(10.1.1)

A

In the one-dimensional model, we assume the velocity to be equal to V everywhere at a given cross section. This model provides excellent results and is used widely. Flows in open channels are most likely to be turbulent, and the velocity profile can be assumed to be approximately constant, as in Fig. 10.1c, without significant error. Hence the one-dimensional model is employed.

10.2 OPEN-CHANNEL FLOWS 10.2.1

Classification of Free-Surface Flows

Flow in a channel is characterized by the mean velocity, even though a velocity profile exists at a given section, as shown in Fig. 10.1. The flow is classified as a combination of steady or unsteady, and uniform or nonuniform. Steady flow signifies that the mean velocity V, as well as the depth y, is independent of time, whereas unsteady flow necessitates that time be considered as an independent variable. Uniform flow implies that V and y are independent of the position coordinate in the direction of flow; nonuniform flow signifies that V and y vary in magnitude along that coordinate. The possible combinations are shown in Table 10.1; the position coordinate is designated as x. Steady, uniform flow is the situation where terminal velocity has been reached in a channel of constant cross section; not only is the mean velocity constant, but the depth is invariant as well. Steady, nonuniform flow is a common occurrence in rivers and in man-made channels. In those situations, it will be found to occur in two ways. In relatively short reaches, called transitions, there is a rapid change

KEY CONCEPT In relatively short reaches, called transitions, there is a rapid change in depth and velocity.

Table 10.1 Combinations of One-Dimensional Free-Surface Flows Type of flow Steady, uniform Steady, nonuniform Unsteady, uniform Unsteady, nonuniform

Average velocity V const. V V(x) V V(t) V V(x, t)

Depth y const. y y(x) y y(t) y y(x, t)


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Rapidly varied flow: A rapid change in depth and velocity.

Gradually varied flow: In extensive reaches of a channel, the velocity and depth change in a slow manner.

in depth and velocity; such flow is termed rapidly varied flow. Examples are the hydraulic jump (shown in Example 4.12), flow entering a steep channel from a lake or reservoir, flow close to a free outfall from a channel, and flow in the vicinity of an obstruction such as a bridge pier or a sluice gate. Along more extensive reaches of channel the velocity and depth may not vary rapidly, but rather, change in a slow manner. Here the water surface can be considered continuous, and the regime is called gradually varied flow. Examples of gradually varied steady flow are the backwater created by a dam placed in a river and the drawdown of a water surface as flow approaches a falls. Figure 10.2 illustrates how both rapidly varied flow (RVF) and gradually varied flow (GVF) can occur simultaneously in a reach of channel. Note that the vertical scale is larger than the horizontal scale; such scale distortion is common when representing open-channel flow situations. Unsteady uniform flow rarely occurs, but unsteady nonuniform flow is common. Flood waves in rivers, hydraulic bores, and regulated flow in canals are all examples of the latter category. In many situations, these flows may be considered to behave sufficiently like steady uniform flow or steady nonuniform flow to justify treating them as such. Unsteady flows are beyond the scope of a fundamental treatment and will not be presented in this book; the one exception is the hydraulic bore—a moving hydraulic jump—which can be analyzed in a quasisteady fashion.

10.2.2 Significance of Froude Number The primary mechanism for sustaining flow in an open channel is gravitational force. For example, the elevation difference between two reservoirs will cause water to flow through a canal that connects them. The parameter that represents this gravitational effect is the Froude number,

Hydraulic jump

Sluice gate

Hydraulic jump

RVF

GVF

Fig. 10.2

RVF

GVF

RVF

GVF

RVF

Steady nonuniform flow in a channel.


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Sec. 10.2 / Open-Channel Flows 477

V Fr 兹gL 苶

(10.2.1)

which has been observed in Chapter 6 to be the ratio of inertial force to gravity force. In the context of open-channel flow, V is the mean cross-sectional velocity, and L is a representative length. For a channel of rectangular cross section, L is the depth y of the flow. The Froude number plays the dominant role in open-channel flow analysis. It appears in a number of relations that will be developed later in this chapter. Furthermore, by knowing its magnitude, one can ascertain significant characteristics regarding the flow regime. For example, if Fr 1, the flow possesses a relatively high velocity and shallow depth; on the other hand, when Fr 1, the velocity is relatively low and the depth is relatively deep. Except in the vicinity of rapids, cascades, and waterfalls, most rivers experience a Froude number less than unity. Constructed channels may be designed for Froude numbers to be greater or less than unity, or to vary from greater than unity to less than unity along the length of the channel.

10.2.3 Hydrostatic Pressure Distribution Consider a channel in which the flow is nearly horizontal, as shown in Fig. 10.3. In this case, there is little or no vertical acceleration of fluid within the reach, and the streamlines remain nearly parallel. Such a condition is common in many open-channel flows, and indeed, if slight variations exist, the streamlines are assumed to behave as if they are parallel. Since the vertical accelerations are nearly zero, one can conclude that in the vertical direction the pressure distribution is hydrostatic. As a result, the sum ( p gz) remains a constant at any depth, and the hydraulic grade line coincides with the water surface. It is customary in open-channel flow to designate z as the elevation of the channel bottom and y as the depth of flow. Since at the channel bottom p/g y, the hydraulic grade line

Hydrostatic pressure distribution assumed

V2 ––– 2g Energy grade line (EGL) y

θ

Water surface (WS) V

1 S0

z

x Datum

Fig. 10.3

Reach of open-channel flow.


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is given by the sum (y z). The concepts in this chapter are developed assuming a hydrostatic pressure distribution.

10.3 UNIFORM FLOW

KEY CONCEPT Channel cross sections can be considered regular or irregular.

Before we study nonuniform flow in Section 10.6, let us focus our attention on the simpler condition of steady, uniform flow or simply uniform flow. Such a flow is rare, but if it were to occur in a channel, the depth and velocity would not vary along its length, or in other words, terminal conditions would have been reached. In addition to uniform flow, this section covers the cross-sectional geometry of open channels; the formulations will apply to both uniform and nonuniform flow. Some of that material was covered in Section 7.7; it is reviewed here for completeness.

10.3.1 Channel Geometry Regular section: Section whose shape does not vary along the length of the channel.

Channel cross sections can be considered to be either regular or irregular. A regular section is one whose shape does not vary along the length of the channel, whereas an irregular section will have changes in its geometry. In this chapter we consider mostly regular channel shapes; three common geometries are shown in Fig. 10.4. The simplest channel shape is a rectangular section. The cross-sectional area is given by (10.3.1)

A by

Wetted perimeter: The length of the line of contact between the liquid and the channel.

in which b is the width of the channel bottom (see Fig. 10.4a). Additional parameters of importance for open-channel flow are the wetted perimeter, the hydraulic radius, and the width of the free surface. The wetted perimeter P is the length of the line of contact between the liquid and the channel; for a rectangular channel it is P b 2y

(10.3.2)

B B

B

y

y

1

1 m2

m1 b

b

(a)

(b)

y

α

+

d

(c)

Fig. 10.4 Representative regular cross sections: (a) rectangular; (b) trapezoidal; (c) circular.


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Sec. 10.3 / Uniform Flow 479

The hydraulic radius R is the area divided by the wetted perimeter, that is, A by R P b 2y

Hydraulic radius: The area divided by the wetted perimeter.

(10.3.3)

The free-surface width B is equal to the bottom width b for a rectangular section. A trapezoidal section (Fig. 10.4b) has the added feature of sloped side walls. If m1 is the ratio of the horizontal to vertical change of the wall on one side, and m2 is the corresponding quantity on the other wall, the area, wetted perimeter, and free-surface width are given as 1 2

A by y2(m1 m2)

(10.3.4)

P b y( 兹苶 1 m苶21 兹苶 1 m苶22 )

(10.3.5)

B b y(m1 m2)

(10.3.6)

Note that the rectangular section is embodied in the trapezoidal definition, since for vertical side walls m1 and m2 are zero, and the relations for A, P, and B become identical to those for the rectangular channel. In addition, if b is set equal to zero, Eqs. 10.3.4 to 10.3.6 describe the geometry of a triangular-shaped channel. The circular cross section is an important one to consider, since many freesurface flows in drainage and sewer systems are conveyed in circular conduits. If d is the conduit diameter, the area, wetted perimeter, and free-surface width are given by d2 A (a sin a cos a) 4

(10.3.7)

P ad

(10.3.8)

B d sin a

(10.3.9)

where y a cos1 1 2 d

冢

冣

(10.3.10)

The angle a is defined in Fig. 10.4c. A generalized cross-sectional geometry can be expressed in functional form as A(y), P(y), R(y), and B(y). The functional representations encompass all of the analytical forms given above, and they also may be used to describe an irregular channel. For example, at a river section, one can describe the area and wetted perimeter in tabular form and utilize techniques such as curve


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Chapter 10 / Flow in Open Channels Flood plain Main channel

(a)

(b)

Fig. 10.5 Generalized section representation: (a) actual cross section; (b) composite cross section.

fitting or interpolation to extract the numerical information as functions of the depth. Such procedures are useful for computer-based analyses. A composite section is one made up of several subsections; usually these subsections are of analytic form. The example shown in Fig. 10.5a consists of a main channel and a floodplain. The main channel is approximated by a trapezoid and the floodplain by a rectangle, Fig. 10.5b. One could derive analytical expressions for such a composite section; however, it may be more useful to consider the functional forms for the geometric parameters. Note that the functions will be discontinuous at depths where the two sections are matched. Most of the theoretical developments in this chapter focus on cross sections that are rectangular. Such an assumption allows one to simplify the mathematics associated with open-channel flow analysis. Even though the equations will be simplified relative to more complicated geometries, the physical understanding of the phenomena and conclusions reached will apply to most generalized prismatic cross sections. A clear distinction will be made between rectangular and other types of geometry when various developments and concepts are presented.

KEY CONCEPT Uniform flow occurs in a channel when the depth and velocity do not vary along its length.

10.3.2 Equation for Uniform Flow Uniform flow occurs in a channel when the depth and velocity do not vary along its length, that is, when terminal conditions have been reached in the channel. Under such conditions, the energy grade line, water surface, and channel bottom are all parallel. Uniform flow can be predicted by an equation of the form V C 兹RS 苶0

(10.3.11)

in which S0 is the slope of the channel bottom and C is the Chezy coefficient, which is independent of the Reynolds number since the flow is considered completely turbulent. It has become common engineering practice to relate C to the channel roughness and the hydraulic radius by use of the Manning relation


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c1 C R1앛6 n

(10.3.12)

where c1 1 for SI units and c1 1.49 for English units. Combining Eqs. 10.3.11 and 10.3.12 with the definition of discharge results in the Chezy–Manning equation c1 Q AR2앛3 兹S 苶0 n

(10.3.13)

Values of the Manning coefficient n are given in Table 7.3. The depth associated with uniform flow is designated y0; it is called either uniform depth or normal depth. Uniform flow rarely occurs in rivers because of the irregularity of the geometry. In man-made channels it is not always present, since the presence of controls such as sluice gates, weirs, or outfalls will cause the flow to become gradually varied. It is, however, necessary to determine y0 when analyzing gradually varied flow conditions, since it provides a basis for evaluating the type of water surface that may exist in the channel. The design of gravity flow sewer networks is often based on assuming uniform flow and the use of Eq. 10.3.13, even though much of the time the flow in such systems may be nonuniform. An examination of Eq. 10.3.13 reveals that it can be solved explicitly for Q, n, or S0. Examples 7.19 and 7.20 provide illustrations. Use of a trial-and-error solution or equation solver is necessary when it is required to find y0 with the remaining parameters given.

KEY CONCEPT The depth associated with uniform flow is called either uniform depth or normal depth.

Example 10.1 Water is flowing at a rate of 4.5 m3/s in a trapezoidal channel (Fig. 10.4b) whose bottom width is 2.4 m and side slopes are 1 vertical to 2 horizontal. Compute y0 if n 0.012 and S0 0.0001. Solution Given geometrical data are b 2.4 m and m1 m2 2. Rearrange Eq. 10.3.13, noting that R A/P and c1 1: A5앛3 nQ 2 P 앛3 2S0 Substituting in the known data and trapezoidal geometry, one has 5/3 1 c 2.4y0 y20(2 2) d 2

3 2.4 y0(2 21 2 2) 4 2/3

0.012 4.5 兹0.0001 苶

Solving for y0, either by trial-and-error or by use of computational software, yields y0 1.28 m.


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Example 10.2 Uniform flow occasionally occurs in a 5-m-diameter circular concrete conduit (Fig 10.4c), but the depth of flow can vary. The Manning coefficient is n 0.013, and the channel slope is S0 0.0005. (a) Calculate the discharge for y 3 m, (b) Plot the discharge-depth curve. Solution (a) First, use Eq. 10.3.10 to find the angle α:

α cos–1(12y/d) cos–1(12 3/5) 101.54° or 101.54 π/180 1.772 rad Using Eqs. 10.3.7 and 10.3.8, the area and wetted perimeter are A

d2 52 (a sina cosa) (1.772 sin 101.54 cos 101.54 ) 12.3 m2 4 4

p ad 1.772 5 8.86 m The hydraulic radius is then A 12.3 1.388 m P 8.86 Finally, the discharge, when y 3 m, is R

1 1 Q AR2/3S1/2 12.3 1.3882/3 0.00050.5 26.3 m3/s n 0.013 (b) Mathcad is employed to generate the curve. Note that the solution is generalized, so that any diameter, Manning coefficient, or channel slope can be entered into the algorithm. Equations 10.3.7 and 10.3.8 are used to define the area and wetted perimeter, respectively. Either SI or English units can be employed by properly defining the parameter c1. In this problem a value of 1.0 is used. A MATLAB solution is shown in Appendix E, Fig E.1. Input diameter, Manning coefficient, and channel slope:

d := 5

n := 0.013

S0 := 0.0005

c1 := 1.0

Define geometric functions:

y a(y) := acosa1 2 # b d d2 # (a(y) sin(a(y)) # cos(a(y))) A(y) := 4 P(y) := a(y) # d R(y) :=

A(y) P(y)

Define discharge function, (i.e., Manning’s equation):

Q(y) :=

2 c1 # A(y) # R(y)3 # 2S0 n


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Plot depth versus discharge:

y := 0,0.01..d 5 4 3 y 2 1 0 0

10

20

30

40

50

Q(y)

10.3.3 Most Efficient Section Design of a channel to convey uniform flow typically consists of selecting or specifying the appropriate geometrical cross section provided that Q, n, and S0 are known. Once chosen, an optimum sizing of the cross section can be based on the criteria of minimum resistance to flow. The flow resistance per unit length equals wall shear stress times wetted perimeter. Using a control volume for uniform flow, assume a small slope S0 so that sin ␪ S0; then, by summing forces on t0P gAS0

(10.3.14)

Hence, a least resistance criterion is equivalent to requiring a minimum crosssectional area with respect to the parameters defining the area. Furthermore, we have to satisfy the Chezy–Manning equation. Since Q, n, and S0 are given and R A/P, Eq. 10.3.13 can be written as P cA5앛2

(10.3.15)

in which c is a constant. As an example, consider a rectangular channel with width b and depth y. The best hydraulic cross section is obtained by rewriting Eq. 10.3.15 such that A becomes a function of b only. Therefore, we express P in terms of A and b, using A by and P b 2y: 2A P b b

(10.3.16)

Substitution of this equation into Eq. 10.3.15 yields 2A b cA5앛2 b

(10.3.17)


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Now, differentiate Eq. 10.3.17 with respect to b, remembering that A is dependent on b: 2 dA 2A 5 dA 1 cA3앛2 b db b2 2 db

(10.3.18)

Set dA/db 0, since the objective is to find the value of b that minimizes A. The result is 2A 1 b2

(10.3.19)

b 2y

(10.3.20)

or, using A by,

Thus, if the width of a rectangular channel is twice the depth of the flowing water, the water will flow most efficiently. For a trapezoidal cross section, it is simpler to begin with Eq. 10.3.15, eliminate b, and express P as a function of A, y, and m, where m m1 m2. Then by considering A as a function of m and y, the minimum of A is found by setting the gradient vector of A to zero. The result is m 兹3苶 앛3 or a side slope angle of 60° with the horizontal. The resulting hexagonal shape is a trapezoid that best approximates a semicircle. The evaluation is left as an exercise for the reader. The optimum criterion based on flow resistance cannot always be used, and in fact may be less significant than other design factors. Additional aspects to be considered are the type of excavation, and if the channel is unlined, the slope stability of the side walls and the possibility of erosion of the bed.

10.4 ENERGY CONCEPTS

Total energy: The sum of the vertical distance to the channel bottom measured from a horizontal datum, the depth of flow, and the kinetic energy.

The energy head at any position along the channel is the sum of the vertical distance measured from a horizontal datum z, the depth of flow y, and the kinetic energy head V 2/2g. That sum defines the energy grade line and is termed the total energy H: V2 H z y 2g

(10.4.1)

The kinetic-energy-correction factor associated with the V 2/2g term is assumed to be unity (see Section 4.4.4); this is common practice for most prismatic channels of simple geometry since the velocity profiles are nearly uniform for the turbulent flows involved. The fundamental energy equation, developed in Section 4.4, states that losses will occur for a real fluid between any two sections of the channel, and hence the total energy will not remain constant. The energy balance is given simply by the relation H1 H2 hL

(10.4.2)


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Sec. 10.4 / Energy Concepts

485

in which hL is the head loss. The only manner in which energy can be added to an open-channel flow system is for mechanical pumping or lifting of the liquid to take place. Equation 10.4.2 is applicable for rapidly varied as well as gradually varied flow situations; it will be used in conjunction with the momentum and continuity equations in a number of applications.

10.4.1 Specific Energy It is convenient in open-channel flow to measure the energy relative to the bottom of the channel; it provides a useful means to analyze complex flow situations. Such a measure is termed specific energy and is designated as E: V2 E y 2g

(10.4.3)

Specific energy then is the sum of the flow depth y and kinetic energy head V 2/2g. Rectangular Sections. For a rectangular section, the specific energy can be expressed as a function of the depth y. The specific discharge q is defined as the total discharge divided by the channel width, that is, Q q Vy b

Specific energy: Measurement of energy relative to the bottom of the channel.

Specific discharge: The total discharge divided by the channel width (valid only for a rectangular channel).

(10.4.4)

The specific energy for a rectangular channel can thus be put in the form q2 E y 2 2gy

(10.4.5)

This E–y relation is shown in Fig. 10.6a. We may observe that a specific discharge requires at least a minimum energy; this minimum energy is referred to as critical energy, Ec. The corresponding depth yc is called the critical depth. If the specific energy is greater than Ec, two depths are possible; those depths are referred to as alternate depths. For constant q, Eq. 10.4.5 is a cubic equation in y for a given value of E which is greater than Ec. The two positive solutions of y are the alternate depths.2 Another way to express Eq. 10.4.5 is to consider E constant and vary q. Equation 10.4.5 can be solved for q as q 兹苶 2gy2 (苶 E y)苶

(10.4.6)

Critical depth: Critical depth is the depth for which specific energy is a minimum. Alternate depths: The two depths of flow that are possible for a given specific energy and discharge.

The E-y curve falls between the two asymptotes E y and y 0. Another curve defined by the relation exists for negative y; it is not considered since it has no physical meaning for open-channel flow.

2


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Chapter 10 / Flow in Open Channels y

y Fr < 1 E = y (asymptote) Fr < 1 yc

Alternate depth points yc

Fr = 1

Fr = 1 Fr > 1 Fr > 1 qmax

E

Ec

(a)

q

(b)

Fig. 10.6 Variation of specific energy and specific discharge with depth: (a) E versus y for constant q; (b) q versus y for constant E.

This relation is shown in Fig. 10.6b. This form of the energy relation is useful for analyzing flows in which the specific energy remains nearly constant throughout the transition region; examples are a change in the width of a channel and the variation of depth with discharge at the entrance to a channel. Note that maximum unit discharge, qmax, occurs at critical depth. The critical depth yc can be evaluated from Eq. 10.4.5 by setting the derivative of E with respect to y equal to zero: dE q2 1 3 0 dy gy

(10.4.7)

Since q Vy, the condition of minimum E is V2 1 1 Fr2 0 gy

(10.4.8)

where the Froude number in a rectangular channel is q V Fr 3 兹gy 苶兹gy 苶

(10.4.9)

Thus, from Eq. 10.4.8 the Froude number is equal to unity for minimum energy. Solving for the depth in Eq. 10.4.7 in terms of q, we have

冢冣

q2 y yc g

1앛3

(10.4.10)

This relation gives the critical depth in terms of the specific discharge. Note that at critical flow conditions, Ec can conveniently be expressed by combining Eqs. 10.4.5 and 10.4.10 to eliminate q, resulting in


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3 2

Ec yc

487

(10.4.11)

On the E–y curve, for any depth greater than yc, the flow is relatively slow or tranquil, and Fr 1; such a state is termed subcritical flow. Conversely, for a depth less than critical, the flow is relatively rapid or shooting, with Fr 1, and the regime is one of supercritical flow.The E–y diagram is a representation of the change in specific energy as the depth is varied, given a constant specific discharge. It is possible for q to vary, as when the width of a rectangular section changes in a transition region. As q increases, the E–y curve shifts to the right in Fig. 10.6a. It is left as an exercise for the reader to show that for a given specific energy, maximization of q of Eq. 10.4.6 will produce critical conditions at maximum discharge (see Fig. 10.6b). Generalized Cross Section. For a generalized section, the specific energy is written in terms of the total discharge Q and the cross-sectional area A as Q2 E y 2 2gA

(10.4.12)

The minimum-energy condition is obtained by differentiating Eq. 10.4.12 with respect to y to obtain dE Q 2 dA 1 3 dy gA dy

(10.4.13)

For incremental changes in depth, the corresponding change in area is dA B dy.Thus setting Eq. 10.4.13 to zero, the minimum-energy condition becomes Q 2B 1 0 gA3

(10.4.14)

By analogy to Eq. 10.4.8, the second term in Eq. 10.4.14 is the square of the Froude number; hence Fr

冪莦

V Q2B Q 앛A 3 gA 兹gA앛B 苶兹gA앛B 苶

(10.4.15)

The ratio A/B is termed the hydraulic depth, and its use allows the definition of the Froude number to be generalized. The reader can verify that A/B is equal to y for a rectangular channel. The specific energy diagram of Fig. 10.6 provides a useful means of visualizing the solution to a transition problem. Even though one probably would solve the problem numerically, an assessment of the graphical solution may provide useful physical insight as well as prevent the incorrect root from being selected.


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Example 10.3 Water is flowing in a triangular channel with m1 m2 1.0 at a discharge of Q 3 m3/s. If the water depth is 2.5 m, determine the specific energy, Froude number, hydraulic depth, and alternate depth. Solution Recognizing that b 0, the flow area and top width are computed from Eqs. 10.3.4 and 10.3.6 as follows: 1 A y 2 (m1 m2) 2 1 2.52 (1 1) 6.25 m2 2 B (m1 m2)y (1 1) 2.5 5.0 m Using Eqs. 10.4.12 and 10.4.15, E and Fr are found to be Q2 E y 2 2gA 32 2.5 2 2.51 m 2 9.81 6.25

冪莦 3 5 冪 0.137 莦 9.81 莦 6.25

Fr

Q 2B gA3 2

3

The hydraulic depth is A 6.25 1.25 m B 5.0 The alternate depth is calculated using the energy equation. Recognizing that A y2, we have 32 2.51 y 2 9.81 (y2)2 0.459 y y4 A trial-and-error solution provides y 0.71 m.

10.4.2 Use of the Energy Equation in Transitions As mentioned in Section 10.2.1, a transition is a relatively short reach of channel where the depth and velocity change, creating a nonuniform, rapidly varied flow. The mechanism for such changes in the flow is usually an alteration of one


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489

y 1

2

Fr < 1

h

Flow

y1 y2 yc

Ec

E

h (a)

(b)

Fig. 10.7 Channel constriction: (a) raised channel bottom; (b) specific energy diagram.

or more geometric parameters of the channel.3 Within such regions, the energy equation can be used effectively to analyze the flow in a transition or to aid in the design of a transition. Two applications are given in this section to demonstrate the methodology; other types of transitions can be treated similarly. Channel Constriction. Consider a rectangular channel whose bottom is raised by a distance h over a short region (Fig. 10.7a). The change in depth in the transition can be analyzed by use of the energy equation, and as a first approximation, one can neglect losses. Assume that the specific energy upstream of the transition is known. Recognizing that H E z, Eq. 10.4.2 is applied from location 1 to the end of the transition region, location 2: E1 E2 h

(10.4.16)

The depth y2 at the end of the transition can be visualized by inspection of the E–y diagram (Fig. 10.7b). If the flow at location 1 is subcritical, y1 is located on the upper leg as shown in the specific energy diagram. The magnitude of h is selected to be relatively small such that y2 yc, hence the flow at location 2 is similarly subcritical; however, as h is increased still further, a state of minimum energy is ultimately reached in the transition. The condition of minimum energy is sometimes referred to as a choking condition or as choked flow. Once choked flow occurs, as h increases, the variations in depth and velocity are no longer localized in the vicinity of the transition. Influences may be observed for significant distances both upstream and downstream of the transition. The reader should verify that in a transition region, a narrowing of the channel width will create a situation similar to an elevation of the channel bottom. The most general transition region is one that possesses a change in both width and in bottom elevation.

KEY CONCEPT The condition of choked flow or a choking condition implies that minimum specific energy exists within the transition.

3 One notable exception to this is the hydraulic jump, which requires use of the momentum principle and will be considered in Section 10.5.2.


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Example 10.4 A rectangular channel 3 m wide is conveying water at a depth y1 1.55 m, and velocity V1 1.83 m/s. The flow enters a transition region as shown in Fig. E10.4, in which the bottom elevation is raised by h 0.20 m. Determine the depth and velocity in the transition, and the value of h for choking to occur. y 1

2

Fr < 1

h

Flow

y1 y2 yc

Ec

E

h (a)

(b)

Fig. E10.4

Solution Use Eq. 10.4.4 to find the specific discharge to be q V1y1 1.83 1.55 2.84 m2앛s The Froude number at location 1 is V1 Fr gy1 兹苶 1.83 0.47 兹9.8 苶 1 苶 1.55 which is less than unity. Hence, the flow at location 1 is subcritical. The specific energy at location 1 is found, using Eq. 10.4.3, to be V2 E1 y1 1 2g 1.832 1.55 1.72 m 2 9.81 The specific energy at location 2 is found, using Eq. 10.4.16, to be E2 E1 h 1.72 0.20 1.52 m If E2 Ec, it is possible to find the depth y2. Therefore, Ec is calculated first. From Eqs. 10.4.10 and 10.4.11 the critical conditions are


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冢冣

q2 yc g

1앛3

冢

冣

2.842 9.81

491

1앛3

0.94 m

3yc 0.94 Ec 3 1.41 m 2 2 Hence, since E2 Ec, we can proceed with calculating y2. The depth y2 can be evaluated by substituting known values into Eq. 10.4.5: 2.842 1.52 y2 2 2 9.81 y 2 The solution is y2 1.26 m q ⬖ V2 y2 2.84 2.25 m앛s 1.26 The flow at location 2 is subcritical since there is no way in which the flow can become supercritical in the transition with the given geometry. The value of h for critical flow to appear at location 2 is determined by setting E2 Ec in Eq. 10.4.16: h E1 Ec 1.72 1.40 0.31 m

Channel Entrance with Critical Flow. Consider flow entering a channel from a lake or reservoir over a short rounded crest, as sketched in Fig. 10.8. If the channel slope is steep, the flow will discharge freely into the channel and supercritical flow will occur downstream of the entrance region. Upstream of the crest the flow may be considered as subcritical. Since the flow regime at the crest changes from subcritical to supercritical, the flow at the crest must be critical. To determine the discharge, we assume that rapidly varied flow takes place over the crest in conjunction with the critical flow condition at the crest. An example illustrates the procedure.

yc

Fr 1 ~ =0 y1

z2 Datum 1

Fig. 10.8

2

Outflow from a reservoir with critical flow at the channel entrance.


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Example 10.5 Water flows freely from a reservoir into a trapezoidal channel with bottom width b 5.0 m and side slope parameters m1 m2 2.0. The elevation of the water surface in the reservoir is 2.3 m above the entrance crest. Assuming negligible losses in the transition and a negligible velocity in the reservoir upstream of the entrance, find the critical depth at the transition and the discharge into the channel.

yc

Fr 1 ~ =0 y1

z2 Datum 1

2

Fig. E10.5 Solution The total energy at location 1 in Fig. E10.5 is y1 since the kinetic energy in the reservoir is negligible (V1 ⬇ 0). Equating the total energies at locations 1 and 2 gives y1 E2 z2 Since critical conditions occur at location 2, Eqs. 10.4.12 and 10.4.14 can be combined to eliminate the discharge, with the result A E2 yc 2B Elimination of E2 in the two equations yields the expression 1 byc (m1 m2)y2c 2 A y1 z2 yc yc 23b (m1 m2)yc 4 2B or, with the given data, the expression becomes 1 5yc (2 2)y2c 2 2.3 yc 235 (2 2)yc 4 The relation above is a quadratic in yc. The positive root is chosen, which gives yc 1.70 m Subsequently, one can find that A 14.28 m2 and B 11.80 m. Use Eq. 10.4.14 to find the discharge to be

冪莦 9.8 14.3 冪莦 49.3 m 앛s .8 11莦

Q

gA3 B

3

3


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493

Energy Losses. Energy losses in expansions and contractions are known to be relatively small when the flow is subcritical; however, it may be necessary under certain circumstances that the losses be considered. Equation 10.4.2 includes a loss term that can account for transitional losses. The following experimentally derived formulas can be employed. For a channel expansion use

冢

冣

(10.4.17)

冢

冣

(10.4.18)

V2 V 2 hL Ke 1 2 2g 2g and for a channel contraction use V2 V2 hL Kc 2 1 2g 2g

In Eq. 10.4.17, Ke is an expansion coefficient; it has been suggested (King and Brater, 1963) to use Ke 1.0 for sudden or abrupt expansions, and Ke 0.2 for well-designed or rounded expansions. For the contraction coefficient Kc in Eq. 10.4.18, use Kc 0.5 for sudden contractions, and Kc 0.10 for well-designed contractions. When the flow is supercritical, significant standing-wave patterns can be generated throughout and downstream of the transition region; for such flows, proper design requires that wave mechanics be considered (Chow, 1959; Henderson, 1966).

10.4.3 Flow Measurement The most common means of metering discharge in an open channel is to use a weir. Basically, a weir is a device placed in the channel that forces the flow through an opening or aperture designed to measure the discharge. Specialized weirs have been designed for specific needs; in this section two fundamental types—broad-crested and sharp-crested—will be presented. A properly designed weir will exhibit subcritical flow upstream of the structure, and the flow will converge and accelerate to a critical condition near the top or crest of the weir. As a result, a correlation between discharge and a depth upstream of the weir can be made. The downstream overflow is termed the nappe, which usually discharges freely into the atmosphere. There are a number of factors that affect the performance of a weir; significant among them are the three-dimensional flow pattern, the effects of turbulence, frictional resistance, surface tension, and the amount of ventilation beneath the nappe. The simplified derivations presented here are based on the Bernoulli equation; the other effects can be accounted for by modifying the ideal discharge with a discharge coefficient, Cd. The actual discharge is the ideal discharge multiplied by the discharge coefficient. When possible, it is advantageous to calibrate a particular weir in place in order to obtain the desired accuracy.

Weir: A device placed in a channel that forces the flow through an opening or aperture, often designed to measure the discharge. KEY CONCEPT Flow will converge and accelerate to a critical condition near the crest of the weir.


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Broad-Crested Weir. A broad-crested weir is shown in Fig. 10.9. It has sufficient elevation above the channel bottom to choke the flow, and it is long enough so that the overflow streamlines become parallel, resulting in a hydrostatic pressure distribution. Then, at some position, say location 2, a critical flow condition exists. Assume a rectangular horizontal channel and let the height of the weir be h. Location 1 is a point upstream of the weir where the flow is relatively undisturbed, and Y is the vertical distance from the top of the weir to the free surface at that point.Applying Bernoulli’s equation from location 1 to location 2 at the free surface and neglecting the kinetic energy head at location 1 (V1 ⬇ 0), one has the result V2 h Y h yc c 2g

(10.4.19)

Vc 兹2g(Y 苶苶 yc)

(10.4.20)

Solving for Vc,

For a weir whose breadth normal to the flow is b, the ideal discharge is Q bycVc byc 兹2g(Y 苶苶 yc)

(10.4.21)

Recognizing that Y Ec, Eq. 10.4.11 is used to relate yc to Y, and when substituted into Eq. 10.4.21, the result is

Q

2 2 g bY 3/2 3B 3

(10.4.22)

For a properly rounded upstream edge on the weir, Eq. 10.4.22 is accurate to within several percent of the actual flow; hence, a discharge coefficient is not applied. Sharp-Crested Weir. A sharp-crested weir is a vertical plate placed normal to the flow containing a sharp-edge crest so that the nappe will behave as a free jet. Figure 10.10 shows a rectangular weir with a horizontal crest extending across the entire width of the channel. Because of the presence of the side walls, lateral contractions are not present. Vc2 ––– 2g EGL Nappe

Y

yc

Fr1 ~ =0 h

2

1

Fig. 10.9

Broad-crested weir.


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η

495

Nappe

ν2

Y

h

ν1 1

2

1

(a)

2 (b)

Fig. 10.10 Rectangular sharp-crested weir: (a) ideal flow; (b) actual flow.

Let us define an idealized flow situation: The flow in the vertical plane does not contract as it passes over the crest so that the streamlines are parallel, atmospheric pressure is present in the nappe, and uniform flow exists at location 1 with negligible kinetic energy (√1 ⬇ 0). The Bernoulli equation is applied along a representative streamline (Fig. 10.10a) and solved for √2, the local velocity in the nappe: √2 兹2gh 苶

(10.4.23)

If b is the width of the crest normal to the flow, the ideal discharge is given as

冕

Y

Qb

0

冕

Y

√2 dh b

2 兹2gh 苶 dh b 兹2g 苶 Y 3앛2 3 0

(10.4.24)

Experiments have shown that the magnitude of the exponent is nearly correct but that a discharge coefficient Cd must be applied to accurately predict the real flow, shown in Fig. 10.10b: 2 Q Cd 兹2g 苶 b Y 3앛2 3

(10.4.25)

The discharge coefficient accounts for the effect of contraction, velocity of approach, viscosity, and surface tension. An experimentally (Chow, 1959) obtained formula for Cd has been given as Y Cd 0.61 0.08 h

(10.4.26)

Normally, for small Y/h ratio, Cd 0.61. If the weir crest does not extend to the side walls but allows for end contractions to appear, as in Fig. 10.11a, the effective width of the weir can be approximated by (b 0.2Y).

Fig. 10.11

b

θ

(a)

(b)

Contracted rectangular and V-notch weirs: (a) rectangular; (b) V-notch.


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The V-notch weir (Fig. 10.11b) is more accurate than the rectangular weir for measurement of low discharge. In a manner similar to the development of the rectangular weir relation, the idealized discharge is found by integrating the local velocity throughout the nappe above the crest. A discharge coefficient is applied to yield

冢

冣

8 u Q Cd 兹2g 苶 tan Y 5앛2 15 2

(10.4.27)

For use with water, and for u varying from 22.5° to 120°, experiments (King and Brater, 1963) have shown that a value of Cd 0.58 is acceptable for engineering calculations. Derivation of Eq. 10.4.27 is left as an exercise. The above equations can, as usual, be used with either set of units.

Example 10.6 Determine the discharge of water over a rectangular sharp-crested weir, b 1.25 m, Y 0.35 m, h 1.47 m, with side walls and with end contractions. If a 90° V-notch weir were to replace the rectangular weir, what would be the required Y for a similar discharge? Solution For the rectangular weir, using Eq. 10.4.26, the discharge coefficient is Y 0.35 Cd 0.61 0.08 0.61 0.08 0.63 h 1.47 Substitute into Eq. 10.4.25 and calculate 2 苶 bY 3앛2 Q Cd 兹2g 3 2 0.63 兹2苶81 9.苶 1.25 0.353앛2 3 0.48 m3앛s With end contractions the effective width of the weir is reduced by 0.2Y, resulting in 2 苶 (b 0.2Y) Y 3앛2 Q Cd 兹2g 3 2 0.63 兹2 苶81 9.苶 (1.25 0.2 0.35) 0.353앛2 3 0.45 m3앛s With a discharge of Q 0.48 m3/s, use Eq. 10.4.27 to find Y for the 90° V-notch weir: 2/5

Y £

Q Ca

8 22g tan(u/2) 15

§ 2/5

£

0.482 0.58

8 22 9.81 tan 45 15

§

0.66 m


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Additional Methods of Flow Measurement. Other types of weirs include those whose faces are inclined in the upstream and downstream direction (triangular, trapezoidal, irregular). In addition, the spillway section of a dam can be considered as a weir with a rounded crest. King and Brater (1963) give details on the selection and use of these types. A special type of open flume is one in which the geometry of the throat is constricted in such a manner as to choke the flow, thereby creating critical flow followed by a hydraulic jump. When manufactured using a particular standardized section, the flume is called a Parshall flume, shown in Fig. 10.12. Extensive calibrations have established reliable empirical formulas to predict the discharge. For throat widths from 1 to 8 ft (approximately 0.3 to 2.4 m), the discharge is given by the formula 0.026

Q 4BH1.522B

497

Parshall flume: An open flume where the throat is constricted to choke the flow to create critical flow followed by a hydraulic jump.

(10.4.28)

in which H is the depth measured at the upstream location shown in Fig. 10.12. Note that H and B are measured in feet, and Q is in cubic feet per second. The other dimensions are used to develop Eq. 10.4.28. In a natural section of a river it may be impractical to place a weir; in that case stream gaging can be used to measure the discharge. A control location is established upstream of the gaging site, and for a given depth, or stage, of the river, the two-dimensional velocity profile is measured using current meters.

2 ft

3 ft

1 5

Flow

6 B

B L = –– + 4 ft 2

Gaging well for measurement of depth H 2 ft. R

1

2L ––– 3

Plan

H 9 in. 3 in. Elevation

Fig. 10.12 Parshall flume. (HENDERSON, OPEN CHANNEL FLOW, 1st,©1966. Printed and Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey.)


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Subsequently, the profile is numerically integrated to yield the discharge. A series of such measurements will produce a stage-discharge curve, which can then be employed to estimate the discharge by measuring the river stage.

10.5 MOMENTUM CONCEPTS In the preceding section we have seen how the energy equation is applied to rapidly varied flow situations, and in particular, how it is utilized to analyze flows in transition regions. The momentum equation is also employed to study certain phenomena in those situations. When used in conjunction with the energy and continuity relations, the momentum equation gives the user a concise means of analyzing nearly all significant transition problems, including problems that involve a hydraulic jump.

10.5.1 Momentum Equation Consider the open-channel reach with supercritical flow upstream of a submerged obstacle, as depicted in Fig. 10.13a. It represents a rapidly varied flow with an abrupt change in depth but no change in width. In general, such a change may result from an obstacle in the flow or from a hydraulic jump. The upstream flow is supercritical and the downstream flow is subcritical. Other flow regimes may also be considered; for example, the conditions could be subcritical throughout the entire control volume. Each situation should be approached as a unique formulation. The generalized flow situation may be used to develop the equation of motion for transition regions. The control volume corresponding to Fig. 10.13a is shown in Fig. 10.13b. The pressure distribution is assumed hydrostatic, and the resultant hydrostatic forces are given by gAy苶, where the distance 苶y to the centroid of the cross-sectional area is measured from the free surface. The submerged obstacle (e.g., a rock or a two-dimensional obstacle) imparts a force F on the control volume

Fr1 > 1 V2

V1

(a)

2 1

– γ A2 y2

– γ A1 y1 F (b)

Fig. 10.13

Channel flow over an obstacle: (a) idealized flow; (b) control volume.


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Sec. 10.5 / Momentum Concepts 499

with a direction opposite to the direction of flow. The linear momentum equation, presented in Section 4.5, is applied to the control volume in the x-direction to give gA1 y苶1 gA2 y苶2 F rQ (V2 V1)

(10.5.1)

Note that frictional forces have not been included in Eq. 10.5.1; they are usually quite small relative to the other terms, so they can be ignored. Similarly, gravitational forces in the flow direction are negligible for the small channel slopes considered. Equation 10.5.1 can be rearranged in the form

F M1 M2 g

(10.5.2)

in which M1 and M2 are terms that contain the hydrostatic force and the momentum flux at locations 1 and 2, respectively. The quantity M is called the momentum function, and for a general prismatic section it is given by Q2 M Ay苶 gA

(10.5.3)

For a rectangular section, Ay苶 by 2/2, and the momentum function is

冢

by 2 bq 2 y2 q 2 M b 2 gy 2 gy

冣

(10.5.4)

y M2 Fr < 1 Conjugate depth points yc

Fr = 1 M1 Fr > 1

M'2

Mc

M F –– γ

Fig. 10.14 Variation of the momentum function with depth.


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Conjugate or sequent depths: The two depths of flow that are possible for a given value of the momentum function and discharge.

Equation 10.5.4 is sketched in Fig. 10.14. Two positive roots y exist for a given M and q; they are termed either conjugate or sequent depths. The upper leg of the curve (y yc) applies to subcritical flow and the lower leg (y yc) to supercritical flow. The values of M1 and M2 for the example of Fig. 10.13a are shown, indicating supercritical flow upstream and subcritical flow downstream of the submerged obstacle. The horizontal distance between M1 and M2 is equal to F/g. The downstream flow would be supercritical if no hydraulic jump occurred; the corresponding value of the momentum function is indicated by M 2. The depth associated with a minimum M is found by differentiating M with respect to y in Eq. 10.5.3: dM BQ 2 A 0 dy gA2

(10.5.5)

Note4 that d(Ay苶)/dy A. The condition for minimum M is thus Q2B gA3

(10.5.6)

This result is identical to Eq. 10.4.14. Hence the condition of minimum M is equivalent to that of minimum energy: critical flow with a Froude number equal to unity. Momentum Equation Applied to a Transition Region. Quite often the momentum equation is applied in situations where it is desired to determine the resultant force acting at a specific location, or to find the change in depth or velocity when there is a significant undefined loss across the transition region. It is important to remember that the energy and continuity equations are also at our disposal, and we must determine which relations are necessary. In some instances, along with the continuity equation, both energy and momentum equations must be applied. An example is given below to illustrate the technique.

4 This can be observed from the definition 苶y (1/A) 兰 (y h) dA. Differentiation of (y 苶A) using Leibnitz’s rule from calculus yields

d d (y A) dy 苶 dy

冕

y

(y h)„(h)dh

0

冨

(y h)„(h)

hy

冕

y

„(h)dh

0

0AA

– y

y

dη

η w (η)


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Example 10.7 In a rectangular 5-m-wide channel, water is discharging at 14.0 m3/s (Fig. E10.7). Find the force exerted on the sluice gate when y1 2 m and y2 0.5 m. Fr1 < 1 Q Fr2 > 1

1

2

Fig. E10.7 Solution Using Eq. 10.5.3, the momentum functions at 1 and 2 are Q2 M1 A1y苶1 gA1 (14)2 5 2 1 12.0 m3 9.81 5 2 Q2 M2 A2y苶2 gA2 (14)2 5 0.5 0.25 8.62 m3 9.81 5 0.5 The resultant force acting on the fluid control volume is determined, using Eq. 10.5.2, to be F g(M1 M2) 9800 (12.0 8.62) 33 100 N Hence the force on the gate acts in the downstream sense with a magnitude of 33.1 kN.

10.5.2 Hydraulic Jump A hydraulic jump is a phenomenon in which fluid flowing at a supercritical state will undergo a transition to a subcritical state. Boundary conditions upstream and downstream of the jump will dictate its strength as well as its location. An idealized hydraulic jump is shown in Fig. 10.15. The strength of the jump varies widely, as shown in Table 10.2, with relatively mild disturbances occurring at one extreme, and significant separation and eddy formation taking place at the other. As a result, the energy loss associated with the jump is considered to be unknown, so the energy equation is not used in the initial analysis. Assuming no

Hydraulic jump: A phenomenon where fluid flowing at a supercritical state will undergo a transition to a subcritical state.


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EGL hj Fr1 > 1 V2

V1

Fr2 < 1

1

2

Fig. 10.15 Idealized hydraulic jump.

friction along the bottom and no submerged obstacle; that is, setting F equal to zero, Eq. 10.5.2 shows that M1 M2. For a rectangular section, Eq. 10.5.4 can be substituted into the relation, allowing one to obtain

冢

冣

q2 1 1 1 (y 22 y 21) g y1 y2 2

(10.5.7)

Rearranging, factoring, and noting that Fr21 q2/gy31 there results

冢

冣

1y y Fr21 2 2 1 2 y1 y1

(10.5.8)

This equation is dimensionless and it relates the Froude number upstream of the jump with the ratio of the downstream to upstream depths. It can be seen that Eq. 10.5.8 is quadratic with respect to y2/y1 provided that Fr1 is known. Solving for y2/y1, one obtains y2 1 兹苶 1 8F苶 r 12 1 y1 2

冢

冣

(10.5.9)

The positive sign in front of the radical has been chosen to yield a physically meaningful solution. It is worth noting that Eq. 10.5.9 is also valid if the subscripts on the depths and Froude number are reversed: y1 1 兹苶 1 8F苶 r 22 1 y2 2

冢

冣

(10.5.10)

The theoretical energy loss associated with a hydraulic jump in a rectangular channel can be determined once the depths and flows at locations 1 and 2 are known. The energy equation is applied from 1 to 2, including the head loss hj


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across the jump, as displayed in Fig. 10.15. Combining the energy equation with Eq. 10.5.7 and the continuity equation, one can show after some algebra that (y2 y1)3 hj 4y1y2

(10.5.11)

Equations 10.5.8 through 10.5.11 are useful forms for solving most hydraulic jump problems in a rectangular channel. Table 10.2 shows the various forms that a hydraulic jump may assume relative to the upstream Froude number. A steady, well-established jump, with 4.5 Fr 9.0, is often used as an energy dissipator downstream of a dam or spillway. It is characterized by the existence of breaking waves and rollers accompanied by a submerged jet with significant turbulence and dissipation of energy in the main body of the jump; downstream, the water surface is relatively smooth. For Froude numbers outside the range 4.5 to 9.0, less desirable jumps exist and may create undesirable downstream surface waves. The length of a jump is the distance from the front face to just downstream where smooth water exists; a steady jump has an approximate length of six times the upstream depth. Table 10.2 Hydraulic Jumps in Horizontal Rectangular Channels Upstream Fr

Type

1.0–1.7

Undular

1.7–2.5

Weak

2.5–4.5

Oscillating

4.5–9.0

Steady

Stable and well-balanced; energy dissipation contained in main body of jump

9.0

Strong

Effective, but with rough, wavy surface downstream

Description Ruffled or undular water surface; surface rollers form near Fr 1.7 Prevailing smooth flow; low energy loss

Intermittent jets from bottom to surface, causing persistent downstream waves

Oscillating jet Roller

Source: Adapted with permission from Chow, 1959. (Adapted from Chow, 1959)


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Example 10.8 A hydraulic jump is situated in a 4-m-wide rectangular channel. The discharge in the channel is 7.5 m3/s, and the depth upstream of the jump is 0.20 m. Determine the depth downstream of the jump, the upstream and downstream Froude numbers, and the rate of energy dissipated by the jump. Solution Find the unit discharge and upstream Froude number: Q q b 7.5 1.88 m2앛s 4 q Fr1 3 兹gy 苶 1 1.88 3 6.71 兹苶 9.81 苶 0.20 The downstream depth is computed, using Eq. 10.5.9, to be y1 1 8F苶 r 21 1) y2 (兹苶 2 0.20 (兹苶 18苶 6.712苶 1) 1.80 m 2 The downstream Froude number is q Fr2 兹苶 gy23 1.88 3 0.25 兹苶 9.81 苶 1.80 The head loss in the jump is given by Eq. 10.5.11: ( y2 y1)3 hj 4y1 y2 (1.80 0.20)3 2.84 m 4 0.20 1.80 Hence, the rate of energy dissipation in the jump is (see Eq. 4.5.25) gQhj 9800 7.5 2.84 2.09 105 W KEY CONCEPT A translating hydraulic jump is a positive surge wave, maintaining a stable front as it propagates into an undisturbed region.

or

209 kW

Translating Hydraulic Jump. A translating hydraulic jump, alternately termed a surge or a hydraulic bore, is shown in Fig. 10.16a; it is termed a positive surge wave in the sense that it maintains a stable front as it propagates at speed „ into an undisturbed region. Such a wave can be generated by abruptly


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ω V2

1

V2 + ω

V1 + ω

V1 2

1

2

(a)

(b)

Fig. 10.16 Translating hydraulic jump: (a) front moving upstream; (b) front appears stationary by superposition.

lowering a downstream gate or by rapidly releasing water at an upstream location into a channel. This unsteady rapidly varied flow situation can be conveniently analyzed as a steady-state problem by superposing the bore speed „ in the opposite sense on the control volume (Fig. 10.16b). The front appears stationary, and the relative velocities at locations 1 and 2 are equal to V1 „ and V2 „, respectively. Assume a bore translating in a rectangular, frictionless, horizontal channel. Equation 10.5.9 can be applied by substituting V1 „ for V1 in the definition of Fr1 to obtain y2 1 y1 2

(V „) 18莦 1冥冤冪莦 gy 莦 2

1

(10.5.12)

1

The continuity relation applied to the control volume of Fig. 10.16b is y1(V1 „) y2(V2 „)

(10.5.13)

Equations 10.5.12 and 10.5.13 contain five parameters: y1, y2, V1, V2, and „. Three of them must be known to solve for the remaining two. Depending on which variables are unknown, the solution of Eqs. 10.5.12 and 10.5.13 will be either explicit or based on a trial-and-error procedure.

Example 10.9 Water flows in a rectangular channel with a velocity of 2.5 m/s and a depth of 1.5 m. A gate is suddenly completely closed, forming a surge that travels upstream. Find the speed of the surge and the depth behind the surge. Solution Since the gate is closed, the downstream velocity is V2 0. Equations 10.5.12 and 10.5.13 contain the two unknowns w and y2. Combining them to eliminate y2 and substituting V1 2.5 and y1 1.5 results in the relation „ 2.5 „ 2

(2.5 „) 18 1冥冤冪莦莦 9.81 莦 1.5 2

(continued)


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A trial-and-error solution, yields „ 3.41 m/s. Use Eq. 10.5.13 to compute y2: V1 „ y2 y1 V2 „ 2.5 3.41 1.5 2.60 m 3.41

Drag on Submerged Objects. If an object is submerged in the flow it is possible to describe the drag force in the manner V2 F CD Ar 2

(10.5.14)

in which CD is the drag coefficient and A is the projected area normal to the flow. A discussion of drag coefficients for various immersed objects is given in Section 8.3.1. In open channel flow, since a free surface is present, the drag coefficient must be modified to account for wave drag as well as drag due to friction and separation. Examples of submerged objects in open channel flow include pipelines and bridge piers; in these situations, a hydraulic jump may not be present. A design application is illustrated in Fig. 10.17. Baffle blocks are devices placed in a reach of channel known as a stilling basin to stabilize the location of a hydraulic jump and to aid in the dissipation of flow energy. Typically they are used for Fr1 greater than 4.5. Example 10.10 demonstrates how baffle blocks reduce the magnitude of the momentum function downstream of a hydraulic jump, resulting in reduced downstream depth and increased energy dissipation. In practice, Eq. 10.5.14 is seldom employed to analyze or design a stilling basin, since other factors such as additional appurtenances, approach velocity, scour, and cavitation must also be considered. Instead, design standards have been established based on observations of existing basins and laboratory model studies (U.S. Department of Interior, 1974; Roberson et al., 1988).

Q

b

y1 y2

1

h

2

Fig. 10.17 Stilling basin with baffle blocks.


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Example 10.10 In the flow situation presented in Example 10.8, a series of baffle blocks is placed in the channel as shown in Fig. E10.10. Laboratory experimentation has shown that the arrangement has an effective drag coefficient of 0.25, provided that the blocks are submerged in the flow. If the blocks are 0.15 m high, and if the discharge, upstream depth and width remain the same as in Example 10.8, determine the depth downstream of the jump and the rate of energy dissipated by the jump.

Q

b

y1 y2

1

h

2

Fig. E10.10 Solution It is necessary to use Eq. 10.5.2 since obstacles (i.e., the baffle blocks) are placed within the control volume. The upstream velocity is Q V1 A1 7.5 9.38 m앛s 4 0.2 The force F due to the presence of the baffle blocks is computed using Eq. 10.5.14: V2 F CD Ar 1 2 9.382 0.25 (4 0.15) 1000 6600 N 2 Note that the frontal area is the width of the channel multiplied by the height of the blocks. Substituting known conditions into Eq. 10.5.2, making use of Eq. 10.5.4 which defines M for a rectangular channel, and noting that q 7.5/4 1.88 m2/s, we find

冢

冣冢

冣

y21 y22 F q2 q2 b b 2 2 g gy1 gy2

冢

冣冢

冣

y22 0.22 1.882 1.882 6600 4 4 2 2 9.81 0.2 9.81y2 9800

(continued)


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The relation reduces to 0.721 y 22 3.31 y2 The solution for y2 is 1.70 m. The change in specific energy between locations 1 and 2 is

冢

冣 1.88 1.88 0.2 冢1.70 冣 2 9.81 0.2 2 9.81 1.70

q2 q2 E1 E2 y1 2 y2 2 2gy1 2gy2 2

2

2

2

2.94 m The rate of energy dissipation, therefore, is gQ(E1 E2) 9800 7.5 2.94 2.16 105 W

or

216 kW

10.5.3 Numerical Solution of the Momentum Equation For nonrectangular channels, the momentum relation can be used directly to analyze the hydraulic jump or other problems requiring the momentum equation; the technique is demonstrated as follows. Consider a trapezoidal channel with conditions known at location 1 upstream of the jump. Consequently, M1 and F are evaluated as constants and Eq. 10.5.2 can be written in the form F M2 M1 0 g

(10.5.15)

in which M2 is a function of y2. Introducing the trapezoidal geometry at location 2, with m1 m2 m, the relation above can be written as y22 Q2 F (2my2 3b) 2 M1 0 6 g g(by2 my2)

(10.5.16)

This can be solved for y2 by a suitable numerical technique such as interval halving, false position, or Newton’s method, which are discussed in any book on numerical methods (see, e.g., Chapra and Canale, 1998). Note that by setting F 0, it becomes the relation for finding the conditions downstream of a hydraulic jump, and that by additionally letting m 0, a rectangular hydraulic jump problem can be solved.


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Example 10.11 A hydraulic jump occurs in a triangular channel with m1 m2 2.5. The discharge is 20 m3/s and yc 1.67 m. Upstream of the jump the following parameters are given: y1 0.75 m, Fr1 7.42, and M1 29.35 m3. Determine the conjugate depth y2 downstream of the jump. Solution Use Eq. 10.5.16 with F 0: y22 202 2 2.5y2 2 29.35 0 6 9.81 2.5y2 The relation reduces to 19.58 35.23 0 f( y2) y32 y22 The false position method is chosen to find y2. The first step is to set the upper and lower limits, termed yu and yl. Since Fr1 1, and y2 yc, an appropriate lower limit is yl yc 1.67 m. The upper limit is assumed to be 5 m.

Iteration

yu

yl

f(yu )

f(yl )

yr

f(yr )

Sign of f(yl ) f(yr )

1 2 3 4 5 6 7 8 9

5 5 5 5 5 5 5 5 5

1.67 2.357 2.808 3.038 3.142 3.187 3.205 3.213 3.216

90.55 90.55 90.55 90.55 90.55 90.55 90.55 90.55 90.55

23.55 18.61 10.61 5.067 2.231 0.944 0.393 0.162 0.0672

2.357 2.808 3.038 3.142 3.187 3.205 3.213 3.216 3.218

18.61 10.61 5.076 2.231 0.944 0.393 0.162 0.0672 0.0277

— — — — — — — — —

2.5 103 9.4 104 4.9 104

The solution is tabulated above. In each iteration a new estimate yr of the root is made: yu f ( yl) yl f( yu) y r f ( yl) f ( yu) The product f (yl) f( yr) is formed to determine on which subinterval the root will be found. If f ( yl) f( yr) 0, then yu yr; otherwise, yl yr. It is required that initially f( yu) and f ( yl) be of opposite sign. Iterations continue until a relative error , defined by

冨

y new y rold r old yr

冨

is less than a specified value, which in the example is 0.0005. The result after nine iterations is y2 3.22 m, rounded off to three significant figures. Using Mathcad or MATLAB, one finds the solution to be less time consuming; see Appendix E, Figs. E.2 and E.3.


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10.6 NONUNIFORM GRADUALLY VARIED FLOW The evaluation of many open-channel flow situations must include accurate analyses of relatively long reaches where the depth and velocity may vary but do not exhibit rapid or sudden changes. In the preceding two sections we emphasized nonuniform, rapidly varied flow phenomena that occur over relatively short reaches, or transitions, in open channels. Attention is now focused on nonuniform, gradually varied flow, where the water surface is continuously smooth. One significant difference between the two is that for rapidly varied flow, losses may often be neglected without severe consequences, whereas for gradually varied flow, it is necessary to include losses due to shear stress distributed along the channel length. The shear stress is the primary mechanism that resists the flow. Gradually varied flow is a type of steady, nonuniform flow in which y and V do not exhibit sudden or rapid changes, but instead vary so gradually that the water surface can be considered continuous. As a result, it is possible to develop a differential equation that describes the incremental variation of y with respect to x, the distance along the channel. An analysis of this relation will enable one to predict the various trends that the water surface profile may assume based on the channel geometry, magnitude of the discharge, and the known boundary conditions. Numerical evaluation of the same equation will provide engineering design criteria.

10.6.1 Differential Equation for Gradually Varied Flow A representative nonuniform gradually varied flow is shown in Fig. 10.18. Over the incremental distance x, the depth and velocity are known to change slowly. The slope of the energy grade line is designated as S. In contrast to uniform flow, the slopes of the energy grade line, water surface, and channel bottom are no longer parallel. Since the changes in y and V are gradual, the energy loss over the incremental length x can be represented by the Chezy–Manning equation. This means that Eq. 10.3.13, which is valid for uniform flow, can also be used to evaluate S for a gradually varied flow situation, and that the roughness coefficients presented in Table 7.3 are applicable. Additional assumptions include a regular cross section, small channel slope, hydrostatic pressure distribution, and one-dimensional flow. 2

V1 ––– 2g

hL = SΔx

S 1

2

y1

V1 y2 S0

z1

EGL

V2 ––– 2g V2

1

Δx

z2

Datum

Fig. 10.18 Nonuniform gradually varied flow.


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Sec. 10.6 / Nonuniform Gradually Varied Flow 511

The energy equation is applied from location 1 to location 2, with the loss term hL given by S x. If the total energy at location 2 is expressed as the energy at location 1 plus the incremental change in energy over the distance x, Eq. 10.4.2 becomes dH H1 H2 S x H1 x S x dx

(10.6.1)

Substitute H y z V 2/2g and dz/dx S0 into this relation and rearrange to find

冢

冣

d V2 S S0 y dx 2g

(10.6.2)

The right-hand term is dE/dx, and it is transformed to dy dE dE dy (1 Fr2) dx dx dy dx

(10.6.3)

(Recall from Eqs. 10.4.13 and 10.4.15 that dE/dy 1 Q 2B/gA3 1 Fr 2.) Finally, upon substitution into the energy relation and solving for the slope dy/dx of the water surface one finds that dy S0 S 2 dx 1 Fr

(10.6.4)

where S and Fr depend on x. This is the differential equation for gradually varied flow and is valid for any regular channel shape.

Example 10.12 Using an appropriate control volume for gradually varied flow, show that the slope S of the energy grade line is equivalent to t0/gR. 1 S EGL – γA y

– d – γ A y + –– (γ A y)Δ x dx

γA Δx

τ 0 PΔ x

Δx

θ

x

Fig. E10.12 (continued)


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Solution The control volume is shown in Fig. E10.12. The resultant force acting on the control volume is due to the incremental change in hydrostatic pressure [gd(Ay苶)/dx] x, the component of weight in the x-direction gAsinu x, and the resistance term t0P x. Using the momentum equation ˙ 2x V1x)

Fx m(V with V2x V1x (dV/dx) x results in d dV g (Ay苶) x gA sin u x t0P x rVA x dx dx This relation can be simplified by noting that d(Ay苶) d(Ay苶) dy dy A dx dy dx dx and P A/R. Substitute and divide the equation by gA x, the weight of the control volume, and find that dy V dV t0 sin u dx g dx gR Since sin u ⯝ S0 for small u, the equation above can be rearranged in the form dy V dV t0 S0 dx g dx gR

冢

冣

V2 d y 2g dx

Upon comparison with Eq. 10.6.2, it is seen that the right-hand side is equivalent to S S0, and consequently, t0 S0 S S0 gR or t0 S gR

10.6.2

Water Surface Profiles

It is possible to identify a series of water surface profiles based on an evaluation of Eq. 10.6.4 (Bakhmeteff, 1932). Essential to the development is the determination of normal and critical depths. Note that y0 and yc are uniquely determined once the channel properties and discharge are established. Table 10.3 shows the classification of the water surface profiles. Associated with yc is a critical slope Sc, which is found by substituting yc into the Chezy–Manning equation and solving for the slope. The channel slope can be designated as mild, steep, or critical, depending on whether S0 is less than, greater than, or equal to Sc, respectively. A horizontal slope exists when S0 0 and an adverse slope when S0 0. Inspection of Table 10.3 shows that there are 12 possible profiles. Each profile is classified by a letter/number combination. The letter refers to the channel slope: M for mild, S for steep, C for critical, H for horizontal, and A for adverse.


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The numerical subscript designates the range of y relative to y0 and yc. Flow can occur at depths above or below yc and at depths above or below y0. The variation of y with respect to x for each profile in Table 10.3 can now be found. For a given Q, n, S0, and channel geometry, the analysis reduces to the determination of how S and Fr vary with y. An inspection of the Chezy–Manning equation will reveal that S decreases with increasing y; similarly, the Froude number decreases as y increases. The numerator in Eq. 10.6.4 will assume the following inequalities: (S0 S) 0 for y y0, and (S0 S) 0 for y y0. In addition, the denominator varies in the manner (1 Fr2) 0 for y yc and (1 Fr2) 0 for y yc. With these criteria, the sign of dy/dx can be evaluated. In addition, with the use of Eq. 10.6.2, the sign of dE/dx is revealed.

Table 10.3 Classification of Surface Profiles Channel slope

Mild S0 Sc y0 yc

Steep S0 Sc y0 yc

Critical S0 Sc y0 yc

Profile type

Depth range

Fr

dy dx

dE dx

M1

y y0 yc

1

0

0

M2

y0 y yc

1

0

0

Horizontal asymptote

M1

y0 M2

yc

M3

M3

y0 yc y

1

0

0

S1

y yc y0

1

0

0

S2

yc y y0

1

0

0

yc

S1 S2

y0

S3

S3

yc y0 y

1

0

0

C1

y yc or y0

1

0

0

y0 = yc

C1 C3

C3

yc or y0 y

1

0

0

Horizontal S0 0 y0 씮

H2

y yc

1

0

0

H3

yc y

1

0

0

Adverse S0 0 y0 undefined

A2

y yc

1

0

0

H2 yc H3

A2 yc

A3

yc y

1

0

A3

0


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The boundaries of the profiles can be established as follows 5: 1. As y tends to y0, S tends to S0. Hence dy/dx approaches zero; in other words, the water surface approaches y0 asymptotically. This applies to curves M1, M2, S2, and S3. 2. As y becomes large, the velocity becomes small and S and Fr tend to zero, so that dy/dx approaches S0. Hence the surface approaches a horizontal asymptote; curves M1, S1, and C1 are of this type. 3. As y approaches yc, dy/dx becomes infinite, a limit that is never reached. For supercritical flow, as the M3, H3, and A3 curves approach yc, a hydraulic jump will form and create a discontinuity in the water surface; at the beginning of the curve, rapid acceleration occurs with nonparallel streamlines. When the flow is subcritical (M2, H2, A2), rapid drawdown takes place close to yc, and the streamlines are no longer parallel. For all of these situations, Eq. 10.6.4 is invalid, since the flow is no longer onedimensional. For the profiles shown in Table 10.3, there is no physical significance to the theoretical limit of y approaching zero, since a finite depth is necessary for the existence of flow. It is left as an exercise to show that for the critical slope condition, dy앛dx Sc as y approaches yc from either a C1 or C2 profile, and that dy/dx approaches a horizontal asymptote as y becomes very large.

Example 10.13 By assuming a wide rectangular channel, develop the right-hand side of Eq. 10.6.4 to show how dy/dx varies with y. Solution For a wide rectangular channel, assume that b >> y, so that the wetted perimeter is approximated by P ⯝ b. The hydraulic radius then becomes R A/P ⯝ (by)/b y. Noting that Q qb, the Chezy–Manning equation, used to evaluate S, simplifies to (qbn)2 (qn)2 S 5 앛3 2 (by ) y10앛3 It is assumed that in the Chezy–Manning equation c1 1. For a rectangular section the square of the Froude number (see Eq. 10.4.9) is q2 Fr2 3 gy Substituting into Eq. 10.6.4 gives dy S0 (qn)2/y10앛3 2 1 q /(gy)3 dx

5

Note that x is measured parallel to the channel bottom, and that y is measured vertically from the channel bottom; hence dy/dx and dE /dx are evaluated relative to the channel bottom and not to a horizontal datum.


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Since (qn)2y010앛3 S0 and Fr2c q2/(gy3c ) 1, the relation can be written as 1 ( y0 앛y)10앛3 dy S0 dx 1 ( yc 앛y)3 This equation can be used as an alternative to Eq. 10.6.4 to evaluate the water surface profiles shown in Table 10.3.

10.6.3 Controls and Critical Flow The existence of the various profiles shown in Table 10.3 depend on the boundary conditions that are specified at given locations in the channel. Quite often, a control will define the boundary condition. A control exists when a depthdischarge relationship can be established at a section. The manner in which the control section affects the water surface away from its specific location can be studied by examining flow near the critical state in a rectangular channel. Equation 10.5.12 is the expression for a surge of finite magnitude translating in the upstream direction in a rectangular channel. As y2 approaches y1, the magnitude of the surge becomes infinitesimal; these waves may be generated by the presence of control structures and other transition sections that tend to “disturb” the flow. In Eq. 10.5.12, one can replace y1 and y2 by y, and V1 by V. In addition, let the wave travel in the downstream direction, so that the equation becomes 1 1 2

(V „) 18 冤冪莦莦gy 莦 1冥

KEY CONCEPT A control is a channel feature that establishes a depthdischarge relationship in its vicinity.

2

(10.6.5)

Solving for „ provides „ V 兹gy 苶Vc

(10.6.6)

in which c 兹gy 苶, termed the celerity. The celerity is the speed at which an infinitesimal wave will travel into an undisturbed region, that is, a region with zero 苶. Thus if a disturbance is crevelocity. For a nonrectangular channel, c 兹gA앛B ated at a midstream location in a channel, two infinitesimal waves are generated. One wave front will tend to propagate upstream at the speed V c, and the second will move downstream at the speed V c. These observations are made relative to an observer in a fixed position, that is, to one who is standing on the shore observing the wave motion. In a rectangular channel with critical flow conditions, Fr V/兹gy 苶 1, or V 兹gy 苶 c. Hence, at critical flow, the first wave front generated by the disturbance would not move upstream but would appear stationary and become a so-called standing wave. The opposite front would be swept downstream at the speed 2c. If Fr 1, the first wave would travel upstream, and the opposite wave would travel downstream at a speed less than 2c. For Fr 1 in the channel, since V c, both waves are swept downstream. Thus, since it contains a mechanism that disturbs the flow, a control will influence upstream conditions only when the


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Chapter 10 / Flow in Open Channels M1 y0 yc

M3

Fr < 1

M2

y01 yc Fr > 1

S2 y02

Fr < 1 S01

Fr > 1

S02

(a)

(b)

S2

M2 Fr > 1

yc y0

(c)

y0 yc Fr < 1 (d)

~ 3.5 y = c

Fig. 10.19 Representative controls: (a) sluice gate; (b) change in slope from mild (S01) to steep (S02); (c) entrance to a steep channel; (d) free outfall.

flow is subcritical (Fr 1). Similarly, when the flow is supercritical (Fr 1), the control can influence only the downstream flow conditions. This is illustrated in Figure 10.19a, where a sluice gate is positioned in a channel with subcritical flow upstream and supercritical flow downstream. An M1 profile is generated above the gate, and an M3 profile exists below it. Any movement of the gate would influence the nature of the two profiles; lowering the gate would lengthen their range, and raising it would produce the opposite effect. The profiles shown in Table 10.3 are all influenced by the presence of controls. In Fig. 10.19, several controls are shown that create a variety of profiles. Critical depth often is associated with an effective control. Examples include sluice gates, weirs, dams, and flumes, all of which force critical flow to occur somewhere in the transition region. In addition, a control can be located at a break in channel slope from mild upstream to steep downstream (Fig. 10.19b). The entrance to a steep channel (Fig. 10.19c) is an example of an upstream control, with critical flow occurring at the crest. Critical flow will occur a short distance upstream from a free outfall on a mild slope (Fig. 10.19d). Other controls, not shown in Fig. 10.19, are a channel constriction acting as a choke, and for a mild slope, the existence of uniform flow at some location.

10.6.4 Profile Synthesis Identification of controls and their interaction with the possible profiles is a requirement for successful understanding and correct design and analysis of open-channel flow. Since controls are essentially transition sections, the rapidly varied flow principles presented in Sections 10.4 and 10.5 can be used to determine the necessary depth–discharge relations. Once the controls have been identified, the profiles can be selected and the range of influence of the controls established. As an example, consider the situation shown in Fig. 10.20. Flow enters the steep channel from a reservoir, so that critical flow exists at the channel entrance.


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Sec. 10.6 / Nonuniform Gradually Varied Flow 517 Conjugate depth curve

S2

S1

yc y0

Fig. 10.20 Example of profile synthesis.

Both the magnitude of the discharge and the S 2 profile are influenced by the depth at the entrance; therefore, the depth acts as a control. At the downstream end of the channel, the lower reservoir acts as a control to establish an S1 profile that projects upstream. At some interior location, a hydraulic jump occurs to allow the flow to pass from a supercritical to a subcritical state. The location of the jump can be found by plotting a curve of the depth conjugate to the S 2 profile and finding its point of intersection with the S1 curve. A lowering of the lower reservoir elevation would cause the jump to move downstream and ultimately be swept out of the channel. Increasing the lower reservoir elevation would move the jump upstream; if it were to move into the entrance region, the upstream control would no longer exist. Additional examples of profile synthesis are given in Section 10.7.

Example 10.14 In a rectangular channel, b 3 m, n 0.015, S0 0.0005, and Q 5 m3/s. At the entrance to the channel, flow issues from a sluice gate at a depth of 0.15 m. The channel is sufficiently long that uniform flow conditions are established away from the entrance region, Fig. E10.14a. Find the nature of the water surface profile in the vicinity of the entrance. y0

5 m3/s 0.15 m (a) M3 profile y0 = 1.39 m

yc = 0.66 m

y1 = 0.25 m (b)

Fig. E10.14 Solution First find y0 and yc to determine the type of channel. To find y0, follow the method shown in Example 10.1. Substitute known data into the Chezy–Manning equation: (continued)


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0.015 5 (3y0)5앛3 3.354 兹0.0005 苶 (3 2y0)2앛3 Solving gives y0 1.39 m. Next, the critical depth is computed to be

冢冣 (5앛3) 冤冥 9.81

q2 yc g

1앛3

2 1앛3

0.66 m

Since y0 yc, a mild slope condition exists. The gate is a control and there will be an M3 profile beginning at the entrance, terminated by a hydraulic jump. Downstream of the jump, the condition of uniform flow acts as a control, so at that location the depth is y0, and the Froude number is q Fr0 3 兹gy 苶 0 5앛3 3 0.325 兹苶 9.81 苶 1.39 Using Eq. 10.5.10, the depth before the jump is y0 1 8F苶 r 20 1) y1 (兹苶 2 1.39 18苶 0.32苶 52 1) 0.25 m (兹苶 2 The depths yc and y1 have been calculated to two significant figures, since the Manning coefficient is known to only two significant figures. Figure E10.14b shows the profile.

10.7 NUMERICAL ANALYSIS OF WATER SURFACE PROFILES Section 10.6 dealt with the understanding and interpretation of various aspects of gradually varied flow. An important procedure was outlined: A water surface profile can be synthesized, or predicted, by making use of relevant information about the channel geometry, roughness, and flow, and by determining or assuming the appropriate controls. Once a satisfactory profile synthesis has been conducted, we are in a position to numerically calculate the desired water surface profiles and accompanying energy gradelines. Regardless of the type of method chosen to numerically evaluate gradually varied flow, analysis of a water surface profile on a reach of channel with constant slope usually follows these steps: 1. The channel geometry, channel slope S0, roughness coefficient n, and discharge Q are given or assumed.


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Sec. 10.7 / Numerical Analysis of Water Surface Profiles 519

2. Determine normal depth y0 and critical depth yc. 3. Establish the controls (i.e., the depth of flow) at the upstream and downstream ends of the channel reach. 4. Integrate Eq. 10.6.4 to find y and subsequently E as functions of x, allowing for the possibility of a hydraulic jump to occur within the reach. For a prismatic channel, y0 can be evaluated by applying the Chezy–Manning equation and finding the root of the function Qn 2앛 10 c1AR 3 兹S 苶0

(10.7.1)

Similarly, yc is found from application of the Froude number in the form Q2B 10 gA3

(10.7.2)

Numerical methods, such as the false position or interval halving (Chapra and Canale, 1988), or computational software can be used to solve Eqs. 10.7.1 and 10.7.2. The Chezy–Manning equation, given here in the form of Eq. 10.7.1, is used to represent the depth–discharge variation for nonuniform as well as for uniform flow. Thus one can replace S0 by the slope S of the energy grade line in Eq. 10.7.1 and solve for S in the manner Q 2n2 S(y) c 21 [A(y)]2 [R(y)]4앛3

(10.7.3)

Since Q and n are either given or assumed, the right-hand side of Eq. 10.7.3 is a function of y alone. Two numerical methods are presented in this section to compute water surface profiles along with normal and critical depths. The first, termed the standard step method, is the most widely used, and the second employs an accurate numerical integration scheme. Solutions will be illustrated using Excel, Mathcad, and MATLAB software. The third method is analytical, and uses Eq. 10.6.4 in integrated form, assuming simplified geometric properties of the channel cross section.

10.7.1 Standard Step Method To develop a numerical procedure for solving gradually varied flow problems, we use Eq. 10.6.2; it is applied over the reach of channel shown in Fig. 10.21, resulting in


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EGL Ei

yi

yi + 1

Ei + 1

WS

xi Δx i

xi + 1

Fig. 10.21 Notation for computing gradually varied flow.

冢

冣

dE d V2 y dx dx 2g S0 S( y)

(10.7.4)

For small changes in y and V between xi and xi 1, Eq. 10.7.4 can be approximated by Ei1 Ei

冕

x

i1

[S0 S( y)] dx

x

i

⯝ (xi1 xi)[S0 S( ym)]

(10.7.5)

in which ym ( yi 1 yi)/2. Equation 10.7.5 is solved for xi 1 to yield Ei1 Ei xi1 xi S0 S( ym)

(10.7.6)

The calculations take place stepwise, beginning at a control point or other location where the depth is known. Assume that it is desired to generate a water surface profile along the channel, and compute values of xi, yi, and Ei, for i 1, . . . , k, where k is the location at the opposite end of the reach. Other than xk, it is not necessary that location xi be fixed, so it can take on any value. Beginning at location i, the evaluation of conditions at location i 1 proceeds as follows: 1. Choose yi 1. 2. Compute Ei 1 from Eq. 10.4.12, ym knowing yi and yi 1, and S( ym) from Eq. 10.7.3. 3. Compute xi 1 from Eq. 10.7.6. 4. At location k, trial values of yk are assumed until Eq. 10.7.6 is satisfied for the known value of xk. The standard step method can be carried out using spreadsheet analysis.


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Example 10.15 Water is flowing at Q 22 m3/s in a long trapezoidal channel, b 7.5 m, m1 m2 2.5. A free overfall is located at the downstream end of the channel, where x 2000 m. For n 0.015, S0 0.0006, find the water surface profile and energy grade line for a distance of approximately 800 m upstream from the free outfall. Solution Equations 10.7.1 and 10.7.2 are used to evaluate y0 and yc by substituting in known data: 22 0.015 c 7.5 2y0 21 (2.5)2 d

2/3

10

5/3 1 c 7.5y0 y 20(2.5 2.5) d 20.0006 2

(22)2 3 7.5 yc(2.5 2.5)4

3 1 9.8 c 7.5yc y2c (2.5 2.5) d 2

10

The roots of these equations can be found using a routine such as Excel Solver®; the solutions are y0 1.29 m and yc 0.86 m. Hence the channel is a mild type, and control will be close to the free overfall at the downstream end of the channel. Without any serious loss of accuracy, one can assume that critical conditions will exist at the free overfall. Referring to Table 10.3, the profile upstream of the overfall will be of type M2. An Excel spreadsheet solution is shown in Table E10.15. The upper part shows the values of critical and normal depths found by using Solver. In the residual column are very small numbers that should be close to zero; see the two above equations. The lower part of the table shows the step method solution. Calculations proceed from station 1 to station 5 in a straightforward manner, with arbitrary values of depth selected and placed in the y column. The beginning value of x (2000 m) is placed in the first cell of the x column, and the remaining distances are computed as explained on the previous page. At station 6, different values of y are chosen until the distance is close to the desired value of 1200 m; a depth of 1.27 m results in a distance of 1230 m, which is acceptable. The spreadsheet equations for computing normal and critical depths, and for the step method, are provided in Appendix E. In addition, a MATLAB solution to this problem is shown in Appendix E, Fig. E.5. Table E10.15 Critical

Station 1 2 3 4 5 6

Depth [m] Residual 0.865 1.087E-06

Normal

1.292

1.812E-06

y [m] 0.865 0.950 1.050 1.150 1.250 1.270

A [m2] 8.358 9.381 10.631 11.931 13.281 13.557

V [m/s] 2.632 2.345 2.069 1.844 1.656 1.623

E [m] 1.218 1.230 1.268 1.323 1.390 1.404

ym [m]

S(ym)

¢x 3 m 4

0.908 1.000 1.100 1.200 1.260

2.165E-03 1.527E-03 1.081E-03 7.866E-04 6.574E-04

8 41 114 357 250

x [m] 2000 1992 1951 1837 1480 1230


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Example 10.16 A trapezoidal channel 300 m long conveys water flowing at Q 25 m3/s. The inverse side slopes of the cross section are m1 m2 2.5, the bottom width is 5 m, the bottom slope is S0 0.005, and the Manning coefficient is n 0.013. Compute the water surface profile and energy grade line if the upstream depth yu 0.55 m and the downstream depth yd 2.15 m. Energy grade line

Hydraulic jump Vertical scale

Water surface 1.0 m yu Conjugate depth yc

y0

yd

300 m

Fig. E10.16 Solution An Excel spreadsheet solution using the step method is shown in Table E10.16. First, input data is entered, including the gravitational constant and the corresponding constant c1 in the Manning equation. Since the two inverse side slopes are equal, a single value m is entered. Then critical and normal depths are evaluated using Excel Solver. Since yc y0, the slope is steep. One control point is at the upstream end, and an S3 profile exists in the upper reach of the channel. The profile is computed from station 1 beginning with yu to station 7 where the depth is 0.85 m and the distance is 288 m. Then an S1 profile is calculated from station 9 at the end of the channel beginning with yd, and terminating when y reaches critical depth. A hydraulic jump is located approximately 190 m downstream from the channel entrance. It is located by plotting the curve of depth conjugate to the S3 profile and finding its intersection with the S1 profile as shown in Fig. E10.16. The conjugate depths ycj are computed using Excel Solver.


2.150 2.050 1.950 1.850 1.750 1.650 1.550 1.450 1.350 1.250 1.124

9 10 11 12 13 14 15 16 17 18 19

yc = y0 = y 0.550 0.600 0.650 0.700 0.750 0.800 0.850

25 5

n= m=

1.121 1.204 1.298 1.404 1.524 1.660 1.817 1.999 2.211 2.462 2.848

V 7.130 6.410 5.806 5.291 4.848 4.464 4.128

Residual 2.26E-07 4.363E-07

2.214 2.124 2.036 1.950 1.868 1.791 1.718 1.654 1.599 1.559 1.537

E 3.141 2.694 2.368 2.127 1.948 1.816 1.719

S0 = 0.005 yu = 0.55

2.100 2.000 1.900 1.800 1.700 1.600 1.500 1.400 1.300 1.187

1.575E-04 1.924E-04 2.371E-04 2.951E-04 3.713E-04 4.728E-04 6.102E-04 7.998E-04 1.067E-03 1.514E-03

2.189E-02 1.624E-02 1.231E-02 9.500E-03 7.449E-03 5.923E-03

0.575 0.625 0.675 0.725 0.775 0.825

g= c1 =

S(ym)

300 2.15

ym

L= yd =

19 18 18 17 17 16 15 13 10 6

26 29 33 40 54 105

¢x

9.81 1.00

300 281 263 245 228 211 195 180 167 157 151

26 56 88 128 182 288

x

1.901 1.802 1.711 1.628 1.550 1.477

ycj

8.81E-07 5.19E-07 1.68E-07 9.30E-07 3.85E-07 5.07E-07

Residual

2:19 AM

22.306 20.756 19.256 17.806 16.406 15.056 13.756 12.506 11.306 10.156 8.778

A 3.506 3.900 4.306 4.725 5.156 5.600 6.056

Depth 1.124 0.864

0.013 2.5

12/1/10

Station 1 2 3 4 5 6 7

Q= b=

Table E10.16

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10.7.2 Numerical Integration Method There are a number of numerical methods available to solve Eq. 10.6.4 for channels with regular cross sections, any of which would provide a solution with sufficient accuracy for design and analysis purposes. Worthy of mention are integration techniques using trapezoidal or Simpson’s rules, and solution of the differential form using a Runge–Kutta procedure. The underlying theory of these methods and algorithms for their implementation can be found in texts on numerical methods. McBean and Perkins (1975) discuss convergence problems associated with use of the trapezoidal rule applied to gradually varied flow. A useful integration scheme is the two-point Gauss–Legendre quadrature (Chapra and Canale, 1998). It is particularly well suited for computer use and is usually more accurate than either the trapezoidal or Simpson’s rule methods. Equation 10.6.4 is written in the integral form

xi1 xi

冕冕

y

i1

y

xi

i y

1 Fr2 dy S0 S

i1

G(y) dy

(10.7.7)

y

i

The last integral is approximated by the Gauss–Legendre formula, giving the result

冕

y

y

i1

i

冤冢

yi1 yi yi1 yi 兹3苶앛3(yi1 yi) G(y) dy G 2 2

冢

冣

yi1 yi 兹3苶앛3(yi1 yi) G 2

冣冥

(10.7.8)

The following example illustrates use of the algorithm.


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Example 10.17 Rework Example 10.15 using Gauss–Legendre quadrature. Solution A Mathcad solution is shown below. After the known data is entered and functions are defined, normal and critical depths are computed. Subsequently, the depth is iterated to find corresponding values of distance using Eqs. 10.7.7 and 10.7.8. Input data: Q : 22

M : 2.5

S0 : 0.0006

b : 7.5

n : 0.015

L : 2000

g : 9.81

Define functions: A(y) : b # yM # y2 S(y) :

B(y) : b2 # M # y

P(y) : b2 # y # 21M2

Q2 # n2 Q2 # B(y) 1 Fr(y)2 Fr(y) : G(y) : B g # A(y)3 S0 S(y) A(y)3.3333 # P(y) 1.3333 E(y) : y

Q2 2 # g # A(y)2

Compute normal and critical depths: yn : root(S(y) S0, y, 0.01,5)

yn 1.292

yc : root(Fr(y) 1, y, 0.01,5)

yc 0.865

Compute S2 profile from upstream location:

X0 : 2000

y0 : yc

⌬y : 0.10

N:4

i : 1,2 . . N yi : yi1 ⌬y

xi : xi1

⌬y 2

# ±G±

yi yi1 2

23 # ⌬y 3

≤ G±

yi yi1

23 # ⌬y 3

2

≤≤

Computed depth, distance and specific energy:

0.865 0.965 y ± 1.065 ≤ 1.165 1.265

2 103 1.988 103 x ± 1.938 103 ≤ 1.798 103 1.257 103

1.218 1.235 E(y) ± 1.275 ≤ 1.333 1.4


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10.7.3 Irregular Channels The method demonstrated in Example 10.15 works well with regular channels. When calculating the profile for a reach of river channel in which the crosssectional data are irregular and usually given at fixed locations, a trial solution is necessary at each location. Henderson (1966) outlines a method of computation that includes corrections for composite sections and eddy losses that occur at bends and expansions. For a composite section, the slope of the energy grade line is given by Q2 S 2 ( Ki)

(10.7.9)

in which Ki is termed the conveyance of subsection i:

冢

冣

c1AR2앛3 K i n

(10.7.10)

i

A kinetic-energy coefficient must be applied to the kinetic-energy term; for a composite section use ( Ai)2 a 3 ( Ki)

冘冢KA冣 3 i 2 i

(10.7.11)

With such complexity, it is most convenient to use the step method procedure in computer-coded form. The United States Army Corps of Engineers has developed the algorithm “HEC-RAS River Analysis System” for use in natural channels. The solution is based on the standard step method and is available for use on a personal computer. The program and documentation can be downloaded from the Website www.hec.usace.army.mil/software/hec-ras/.

10.7.4 Direct Integration Methods Equation 10.6.4 can be integrated in a straightforward manner if one assumes a wide rectangular, horizontal channel. A more general integration procedure is available for nonhorizontal prismatic channels of various shapes (Chow, 1959). Assume that in the Chezy–Manning equation the product A2R4/3 is proportional to yN, and in the Froude number the ratio A3/B is proportional to yM, where M and N are constants. Then Eq. 10.6.4 can be written 1 1 (yc앛y)M dx dy S0 1 (y0앛y)N

冤

冥

(10.7.12)

The solution of Eq. 10.7.12 is


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冤

冢冣

冥

M

y0 yc x u F(u, N) S0 y0

J F(√, J) N

(10.7.13)

in which u y앛y0, √ uN/J, J N/(N M 1), and F is the varied flow function

冕

u

F(u, N)

0

dh 1 hN

(10.7.14)

Equation 10.7.14 can be evaluated numerically and the results are presented as the varied flow function in Appendix E, Table E.1. Values of N 2 12, 3, and 3 13 are commonly used since they are consistent with a wide rectangular channel assumption (see Example 10.13). Note that the function F may have any desired constant of integration assigned to it. In the varied flow function table the constant is adjusted so that F(0, N) F(, N) 0. Since integration from one location to another is direct, no intermediate calculations are necessary. This method of evaluation is in direct contrast to the numerical methods, where intermediate steps cannot be avoided without the loss of precision.

Example 10.18 A wide rectangular channel conveys a discharge of q 3.72 m3/s per meter width on a slope of S0 0.001. At a given location the depth is 3 m. Determine the distance upstream where the depth is 2.5 m. The Manning coefficient is 0.025. Solution From Example 10.13, for a wide rectangular channel we found that N 3.33 and M 3. Therefore, one can calculate J to be N 3.33 J 2.5 N M 1 3.33 3 1 Also from Example 10.13, we can determine y0 in the manner

冢冣

q2n2 y0 S0

3앛10

冢

3.722 0.0252 0.001

冣

3앛10

1.91 m

Furthermore,

冢冣

q2 yc g

1앛3

冢

冣

3.722 9.81

1앛3

1.12 m

Substitute these values into Eq. 10.7.13 and simplify:

冤

冢冣

1.91 1.12 x u F(u, 3.33) 0.001 1.91

3

冥

2.5 F(√, 2.5) 3.33

1910[u F(u, 3.33) 0.151F(√, 2.5)] Since y0 yc, and at the downstream location y y0, the profile is an M1 curve. At the downstream location where the depth y 3 m,

(continued)


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y 3 u 1.57 y0 1.91

and

√ uN/J 1.573.33앛2.5 1.825

From Appendix E, Table E.1, making use of linear interpolation between recorded values we find that F(1.57, 3.33) 0.166

and F(1.825, 2.5) 0.300

Considering x as a distance measured from an arbitrary datum, we find that x 1910(1.57 0.166 0.151 0.300) 2763 m To determine the distance upstream of the weir where the depth is 2.5 m, we perform the following calculations in a manner analogous to those at the downstream location: 2.5 u 1.31 1.91

and √ 1.313.33앛2.5 1.43

F(1.31, 3.33) 0.578

and F(1.43, 2.5) 0.474

x 1910(1.31 0.578 0.151 0.474) 1536 m Hence, the distance between the two locations, from a depth of 3 m to where the depth is 2.5 m, is 2763 1536 1227 m, or approximately 1230 m.

10.8 SUMMARY As opposed to pipe flows, open channel flow is complicated by the presence of a free surface. In addition to velocity, the depth of flow is variable, so that even though it may be steady, the flow in a channel is generally nonuniform. In Section 10.3, as well in Chapter 7, we have introduced uniform flow as a limiting case— one that does not often occur but nonetheless is significant. Friction losses in open channel flow are described mathematically by the Chezy–Manning equation, which relates velocity or discharge to the geometric and hydraulic properties of the channel cross section. Energy and momentum principles were introduced in Sections 10.4 and 10.5 and applied to rapidly varied nonuniform flow that occurs at channel transitions and at a hydraulic jump. Critical and conjugate depths have been derived along with the definition of the Froude number. We have seen that specific energy is a useful concept for dealing with transition analysis. Much of the development in this chapter has used a rectangular cross section; this was done primarily for mathematical expediency. The guidelines that were developed and conclusions made using the rectangular section apply to nonrectangular sections as well. Table 10.4 summarizes several of the formulas developed in the chapter. Gradually varied nonuniform flow occurs where the depth and velocity vary continuously with distance along a channel. In Section 10.6, application of the energy equation to gradually varied flow resulted in a mathematical description of the water surface profile. We have presented a classification of the various types of flow regimes that occur along a channel reach. In addition we have shown how the understanding of rapidly varied and gradually varied flow enable


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Problems 529

one to predict the manner in which water surfaces will behave, a task that is required prior to numerical computation of water surface profiles. Several numerical methods were introduced in Section 10.7 to integrate the gradually varied flow equation, and examples of solutions were provided using spreadsheets, computational software, and approximate integrals. Perhaps the most useful means of solution is the spreadsheet, in which a generalized algorithm can be readily adapted for use in different problems. Table 10.4 Formulas for Rectangular and General Sections Section

Fr

Rectangular

General

q 兹苶 gy3

Q/A 兹gA/B 苶

yc

冢qg冣

2 1/3

Q2B 1 gA3

E

M

q2 y 2 2gy

by2 q2 2 gy

Q2 y 2 2gA

Q2 Ay苶 gA

PROBLEMS 10.1

10.2

Give an example of each of the following types of flow for both closed-conduit and free-surface conditions. Include a sketch with each representation: (a) Steady, uniform (b) Unsteady, nonuniform (c) Steady, nonuniform (d) Unsteady, uniform Water is flowing with a free surface at 4 m3/s in a circular conduit. Find the diameter d such that the critical depth yc 0.3d. Under critical flow conditions, Q2B/(gA3) 1.

10.3

Classify the following flows as steady or unsteady, and uniform or nonuniform: (a) Water flowing through a logjam in a river with the observer standing on the shore. (b) Water flowing through a river rapids with the observer on a raft traveling with the current. (c) A flood wave traveling in a river which is generated by a gentle rainfall/runoff event, with the observer standing on the bank. (d) A moving hydraulic jump in a prismatic channel, with the observer running alongside the bank at the same speed as the jump.

Uniform Flow 10.4

For each regular channel section, construct the curves R versus y and AR2/3 versus y: y

(a)

Circular, d 2 m

(b) Trapezoidal, b 3 m, m1 m2 2.5 (c)

Composite (see Fig. P10.4)

2.2 m

1m

20 m

5m

35 m

Fig. P10.4


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10.6

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Chapter 10 / Flow in Open Channels Determine the most efficient section based on flow resistance for a trapezoidal channel. Assume equal side slopes, that is, m1 m2. Determine the uniform depth y0, area A, and wetted perimeter P for the following conditions: (a) Trapezoidal channel, m1 2.5, m2 3.5, b 4.5 m, Q 35 m3/s, n 0.015, S0 0.00035 (b) Circular channel, d 2.1 m, Q 4.25 m3/s, n 0.012, S0 0.001 (c) Circular channel, d 6 ft, Q 120 ft3/sec, n 0.012, S0 0.001 (d) Trapezoidal channel, Q 15 m3/s, n 0.011, S0 0.0013, “most efficient” cross section (e) Rectangular channel, b 25 ft, Q 1200 ft3/sec, n 0.020, S0 0.0004

10.7

10.8

Uniform flow occurs in a trapezoidal channel with m1 m2 1.75. The channel is to convey a discharge of 4.0 m3/s, with an average velocity of 1.4 m/s. If n 0.015 and S0 0.001, find the bottom width and the uniform depth. A channel cross section, commonly called a gutter, forms at the side of a street next to a curb during rainfall conditions (Fig. P10.8). The slope along the roadway is S0 0.0005, and the Manning roughness coefficient is n 0.015. Assuming that uniform flow conditions occur: (a) Determine the discharge if the depth of flow is y0 12 cm (b) If Q 80 L/s, what is the flow depth y0?

1

y0

8

Fig. P10.8 10.9

(a) Given S0 0.0003 and Q 1.75 m3/s, find d (b) Given S0 0.00005 and d 1.3 m, find Q (c) Given d 0.75 m and Q 0.45 m3/s, find S0

A sewer pipe of circular cross section is made of concrete (n 0.014). It is to convey water at uniform flow conditions such that the pipe flows half full, that is, y d/2, where d diameter of the pipe. Evaluate the following conditions: Energy Concepts

10.10

For a given specific energy in a rectangular channel, show that critical flow exists when the specific discharge is at its maximum value qmax. 10.11 The specific energy relation for a rectangular section, Eq. 10.4.5, can be made dimensionless by normalizing it with respect to critical depth yc. Prepare such a plot with y/yc as the ordinate and E/yc as the abscissa. Locate the minimum point mathematically. 10.12 Water is flowing in a rectangular channel at a velocity of 3 m/s and a depth of 2.5 m. Determine the changes in water surface elevation for the following alterations in the channel bottom: (a) An increase (upward step) of 20 cm, neglecting losses. (b) A “well-designed” increase of 15 cm. (c) The maximum increase allowable for the specified upstream flow conditions to remain unchanged, neglecting losses. (d) A “well-designed” decrease (downward step) of 20 cm.

10.13

Water is flowing in a rectangular channel whose width is 5 m. The depth of flow is 2 m and the discharge is 25 m3/s. Determine the changes in depth for the following alterations in the channel width: (a) An increase of 50 cm, neglecting losses (b) A decrease of 25 cm, assuming a “welldesigned” transition

10.14

Flow in a rectangular channel of width b takes place with a known depth y and velocity V. Downstream of this location, there is an upward step of height h. Neglecting losses, determine the change in width that must take place simultaneously for critical flow to occur within the transition. (a) b 3 m, y 3 m, V 3 m/s, h 70 cm (b) b 10 ft, y 10 ft, V 10 ft/sec, h 2.3 ft


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Problems 531 10.15

Q

Water is flowing in a rectangular channel 2.0 m wide. At a transition section the channel bottom is lowered by h 0.1 m for a short distance, and then is raised back to the original elevation (Fig. P10.15). If y 1.22 m and Q 4.8 m3/s, then, with losses neglected: (a) Find the change in channel width necessary to maintain a horizontal water surface through the transition. (b) What change in width would cause critical flow to occur in the transition? y

y

not yet fixed, but the depth y at the entrance and the discharge Q are known. Determine the necessary widths for the following geometries: (a) Rectangular section, y 1 m, Q 18 m3/s (b) Trapezoidal section, m1 m2 3, y 1 m, Q 18 m3/s (c) Rectangular section, y 3 ft, Q 635 ft3/s (d) Trapezoidal section, m1 m2 3, y 3 ft, Q 635 ft3/s 10.19 Rapidly varied flow occurs over a rectangular sill (Fig. P10.19), with free outfall conditions at location C. Throughout the reach losses can be neglected. (a) Compute the discharge. (b) Evaluate the depths at locations A, B, and C. el 103.6 m

h

Fig. P10.15 – el 102.0 m

10.16

Water flows at a depth of 2.15 m and a unit discharge of 5.5 m2/s in a rectangular channel. Energy losses can be neglected. (a) What is the maximum height h of a raised bottom that will permit the flow to pass over it without increasing the upstream depth? (b) Show the solution on an E–y diagram. (c) Sketch the water surface and energy grade line. (d) If the channel bottom is raised greater than h, discuss a type of change that may take place upstream of the transition. 10.17 A lake discharges into a steep channel. At the channel entrance the lake level is 2.5 m above the channel bottom. Neglecting losses, find the discharge for the following geometries: (a) Rectangular section, b 4 m (b) Trapezoidal section, b 3 m, m1 m2 2.5 (c) Circular section, d 3.5 m 10.18 Steep flow conditions exist at the outlet from a reservoir into an open canal. The canal width is

el 100.0 m 2m A

2m B

C

Fig. P10.19 10.20

Prepare a computer algorithm that evaluates the critical depth of a regular channel using the false position technique or a method of your choice. Verify the program with the following data: (a) Circular section, Q 5 m3/s, d 2.5 m (b) Rectangular section, Q 125 ft3/sec, b 12 ft (c) Trapezoidal section, Q 120 m3/s, b 10 m, m1 m2 5 (d) Triangular section, Q 250 ft3/sec, m1 2.5, m2 3.0 10.21 For the channel cross section shown in Fig. P10.21, determine the critical depth if the discharge is 16.5 m3/s.

y

1

1.5 m

3 10 m

Fig. P10.21


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Chapter 10 / Flow in Open Channels 10.23

For the composite channel section shown in Fig. P10.22, find the critical depth yc if: (a) Q 55 m3/s (b) Q 3.5 m3/s

A parabolic channel (Fig. P10.23) is described by the function h 0.1x2. If Q 25 m3/s and y 2.00 m, find the alternate depth. η

dA = 2xdη

b = 3√ y m

dη

y

1m y 10 m

x 10 m

Fig. P10.23

Fig. P10.22

10.24

10.25

Prepare a computer algorithm that determines the alternate depth, given the geometry of a prismatic channel, the discharge, and the depth of flow. Use the interval-halving technique or a solution of your choice. Derive the relation for a V-notch weir that expresses Q as a function of Y, Eq. 10.4.27.

10.26

Debris blocks the entire width of a river as shown in Fig. P10.26. The cross section of the river is approximately rectangular, where b 15 m, Q 22.5 m3/s, y0 1.2 m, and h 0.2 m. (a) Neglecting losses, analyze the rapidly varied water surface in the vicinity of the debris. (b) Sketch the energy grade line and water surface.

y0 Q

h 1

2

3

Fig. P10.26

10.27

Develop the expression for the theoretical discharge for the weir shown in Fig. P10.27. Sharp-crested weir

y

Y

x = y2 x

Fig. P10.27

10.28

A Parshall flume is placed in a small stream to measure the discharge. It has a width of 3 ft at its throat. (a)

Compute the discharge if the measured upstream depth is 1.2 ft.

(b) Prepare a plot of Q versus H where H ranges from 0.5 to 1.5 ft. 10.29 Water is flowing at a given depth and discharge in a 3.5-m-wide rectangular channel. A 60-cm-high submerged raised section is placed in the channel


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Problems 533 as shown in Fig. P10.29. The raised section spans the entire width of the channel, and is designed such that losses can be neglected. Calculate the water surface and energy grade line: (a)

(b) When y 1.0 m and Q 3 m3/s (c) Under what given conditions, (a) or (b) above, would the raised section act similar to a broad crested weir (explain your answer in words only)?

When y 1.5 m and Q 4 m3/s

Q

y1

1

2

Fig. P10.29 10.30

Design a broad-crested weir to convey a river discharge that varies between Q1 and Q2. The maximum water depth upstream of the weir is not to exceed y2 and the minimum depth is not to be less than y1. (a) Q1 0.15 m3/s, Q2 30 m3/s, y1 1.05 m, y2 1.75 m (b) Q1 5 ft3/sec, Q2 1000 ft3/sec, y1 3.45 ft, y2 5.75 ft 10.31 Design a canal to deliver water between the two reservoirs shown in Fig. P10.31. The horizontal distance between the reservoirs is L. It is required that uniform flow exist throughout the reach. The water surface in reservoir A is to be no more than the distance h above the bottom of the canal at the entrance. The canal is rectangular in cross section, and made of concrete. Neglect entrance and

exit losses, and assume uniform, subcritical flow conditions at the channel entrance. (a) ElA 501.8 m, ElB 500.2 m, L 1500 m, h 2 m, b 2.5 m (b) ElA 1646 ft, ElB 1641 ft, L 4920 ft, h 6.5 ft, b 8 ft A B

h(max)

+

+

L

Fig. P10.31

Momentum Concepts 10.32

The momentum function for a rectangular section, Eq. 10.5.4, can be made dimensionless by normalizing it with respect to b(yc)2. Prepare such a plot using y/yc as ordinate and M/(byc)2 as abscissa.

10.33

Rework Problem 10.24 to evaluate the conjugate depth in place of the alternate depth.

10.34

Flow occurs in a wide rectangular channel at a depth y. A construction project over this channel requires that cofferdams spaced at distance „ on centerlines be placed in the channel (Fig. P10.34). Assuming that the cofferdams create no energy losses between locations 1 and 2, determine the maximum permissible diameter of the cofferdams without creating upstream backwater effects (i.e., without increasing the upstream depth), and the

resultant drag force on each cofferdam (assume that CD 0.15). (a) q 1.5 m2/s, y 1.8 m, „ 6 m (b) q 16 ft2/sec, y 6 ft, „ 20 ft

d/2 d/2

q w

Cofferdam 1

2

Fig. P10.34


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A temporary pipeline (diameter 200 mm CD 0.30) is to be placed on a river bed normal to the direction of flow. The river is approximately rectangular in section, and is 100 m wide. The depth of flow downstream of the pipe is 2.5 m and the mean velocity is 3 m/s. Neglecting bed slope and resistance, determine the following: (a) Flow conditions (i.e., depth and velocity) upstream of the pipe once it is in place (b) Resultant drag force on the pipe 10.36 A hydraulic jump occurs over a sill located in a triangular channel with water flowing as shown in Fig. P10.36. The inverse side slopes are m1 m2 3, the drag coefficient CD 0.40, and the height of the sill h 0.3 m. Determine the discharge Q if y1 0.50 m and y2 1.8 m.

(e)

10.35

Q

y1

Describe the nature and character of the jump. h

Q 1

2

3

Fig. P10.39 10.40

y2

Water is flowing as shown in Fig. P10.40 under the sluice gate in a horizontal rectangular channel that is 5 m wide. The depths y1 and y2 are 2.5 m and 10 cm, respectively. The horizontal distances between locations 1, 2, and 3 are sufficiently short that rapidly varied flow conditions can be assumed to occur. Determine the following: (a) The discharge (b) The depth downstream of the jump at location 3 (c) The power lost in the hydraulic jump

h

Fig. P10.36 Q

10.37

Water is flowing in a rectangular channel at a depth of 1.6 m and a velocity of 0.85 m/s. At a downstream location the discharge is suddenly reduced to zero, causing a surge to propagate upstream. Find the depth and velocity behind the surge, and the speed of the surge wave. 10.38 In a rectangular channel, water is flowing at a depth y and a velocity V. At a downstream location, the discharge is suddenly reduced by 60% (i.e., to 40% of the original value), causing a surge to propagate upstream. Determine the depth and velocity behind the surge, and the speed of the surge wave. (a) y 1.5 m, V 1 m/s (b) y 5 ft, V 3 ft/sec 10.39 Water enters a reach of rectangular channel where y1 0.5 m, b 7.5 m, and Q 20 m3/s (Fig. P10.39). It is desired that a hydraulic jump occur upstream (location 2) of the sill and on the sill critical conditions exist (location 3). Other than across the jump, losses can be neglected. Determine the following: (a) Depths at locations 2 and 3 (b) Required height of the sill, h (c) Resultant force acting on the sill (d) Sketch the water surface and energy grade line.

1

2

3

Fig. P10.40 10.41

A transition section is located at the entrance to a rectangular channel as shown in Fig. P10.41. At location 1, the depth is sufficiently large so that the velocity is negligible. If b 3 m, Fr3 0.75, and Q 5.55 m3/s, determine the following: (a) The water surface elevation relative to the datum at locations 1, 2, and 3. Sketch the water surface and energy grade lines between locations 1 and 3. (b) The resultant horizontal force acting on the channel bottom between locations 2 and 3

1 h=–√x 2

+

x

Datum

2m 1

2

3

Fig. P10.41


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Problems 535 Water is flowing in a trapezoidal channel, b 5 m, m1 m2 3. A stationary hydraulic jump occurs, with the upstream depth y1 1.1 m and the discharge Q 60 m3/s. Find the downstream depth y2 and the power dissipated by the jump. 10.43 A 4-m-wide rectangular channel contains an upward rectangular step whose width is the same as the channel. The given flow conditions are y1 0.5 m, V1 8 m/s, and y2 2 m. (a) What is the height h of the step if CD 1.2? (b) What would be y2 if the step were not present? 10.44 A rectangular channel (Fig. P10.44) has a sudden upward step of 0.17 m. A hydraulic jump occurs above the step. At location 2, downstream of the

jump, the depth is y2 1.5 m, and the Froude number there is Fr2 0.40. If the width of the channel is 5 m and the drag coefficient on the step is CD 0.35, find the discharge Q and the depth y1 upstream of the jump.

10.42

y2

Q y1 2 1

0.17 m

Fig. P10.44

Nonuniform Gradually Varied Flow 10.45

The channel whose bottom profile is shown in Fig. P10.45 has a rectangular cross section, with b 8 m, n 0.014, and S0 0.004. Determine the following: (a) The discharge in the channel (b) Sketch the water surface and energy grade line. (c) The possible existence of a hydraulic jump (Hint: Assume critical flow conditions at the channel entrance.)

10.47

Rework Problem 10.46 if S0 0.0013.

10.48

Flow of water takes place in a rectangular channel with b 4 m, n 0.012, L 500 m, S0 0.00087. The discharge is 33 m3/s. At the entrance to the channel, the depth is 0.68 m, and at the downstream end a free-outfall condition exists. Perform a profile synthesis to determine the nature of the water surface and energy grade line.

10.49

A flow of 8.5 m3/s occurs in a long rectangular channel 3 m wide with y0 1.54 m. There is a smooth constriction in the channel to 1.8 m width. (a) Determine the depths to be expected in and just upstream of the constriction, neglecting losses. Show the solution on an Ey diagram. (b) Classify the gradually varied flow profile upstream of the constriction.

10.50

Water is discharging at 20 m3/s in a triangular channel with m1 3.5, m2 2.5, S0 0.001, n 0.014. At the entrance to the channel the depth is 0.50 m, and at a distance 300 m downstream the depth is 2.5 m. Determine the nature of the water surface over the 300-m reach.

10.51

A long channel has a change in bottom slope 500 m from its downstream end; upstream of the transition the slope is S01 0.0003, and downstream of it the slope is S02 0.005. The downstream reach of channel terminates at a reservoir whose elevation is 3 m above the channel bottom at that location. The upper reach is quite long, and

el 103.6 m

el 102 m

el 101.5 m

Fig. P10.45 10.46

Uniform flow occurs in a long rectangular channel of width b, S0 0.0163 and n 0.012. The channel is altered by raising the channel bottom by z for a short distance. The alteration is intended to act as a “smooth transition region” so that losses can be considered negligible. Determine what changes take place due to the presence of the transition. In your analysis include all relevant rapidly varied flow calculations, identify any gradually varied flow profiles, and sketch the water surface and energy grade line. (a) Q 0.35 m3/s, b 1.80 m, z 100 mm (b) Q 12.5 ft3/sec, b 6 ft, z 4 in


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very far upstream from the transition normal flow conditions exist. The channel section is trapezoidal, with a bottom width of 3 m, m1 m2 1.8, n 0.012, and a discharge of 17.5 m3/s. Determine the nature of the water surface and energy grade line from a location far upstream of the transition to the end of the channel. 10.52 A rectangular channel with b 4 m has a change in slope such that the normal depths are y01 0.93 m and y02 1.42 m. At locations far upstream and far downstream of the transition, flow occurs at the respective uniform depths that serve as controls. The discharge is 15 m3/s. Find the variation of the water surface and energy grade line throughout the region.

10.53

The partial water profile shown in Fig. P10.53 is for a rectangular channel of width b 3 m, in which water is flowing at a discharge of Q 5 m3/s. (a) Does a hydraulic jump occur in the channel? If so, is it located upstream or downstream of location A? (b) Sketch the water surface and energy grade line, and identify any known water surface profiles.

1.6 m y0 = 0.4 m (uniform depth)

A

(horizontal slope)

Fig. P10.53

10.54

A very long rectangular channel (Fig. P10.54) has normal flow conditions at locations A and C, and a change in the bottom slope occurs at location B. (a) Somewhere between locations A and C a hydraulic jump will take place. Explain why this statement is true. (b) Classify and sketch the two possible water surface profiles that exist between locations A and C. (c) By appropriate reasoning and calculation, determine whether the jump will occur upstream of location B, or downstream of location B. y01 = 1.0 m

y02 = 1.8 m

q = 5.75 m2/s

y01

10.55

A long rectangular channel (Fig. P10.55) has normal flow conditions at locations A and D, respectively, far upstream and downstream of location C. At C a change in slope occurs as shown. (a) Classify the water surface, and sketch the water surface and the energy grade line between A and D. (b) Determine the height h of a transition such that normal flow conditions occur between A and B, which is a short distance upstream of A. (c) Explain what type of profile now exists downstream of B.

y0 = 1.6 m 1

y02 = 0.8 m

q = 4 m2/s

y01

A y02

A B

y02

C

B D

C

Fig. P10.54

Fig. P10.55


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Problems 537 10.56

Water is flowing at a discharge Q in a circular conduit of diameter d and length L, with S0 0.001, and n 0.015. At the inlet the water depth is y0 and at the outlet a free overfall exists. Determine the nature of the water surface in the conduit. (a) Q 2.5 m3/s, L 500 m, d 2.5 m, y0 0.4 m (b) Q 88 ft3/sec, L 1600 ft, d 8 ft, y0 1.3 ft

10.57

With the results from Problem 10.51, compute the water surface profile and energy grade line using a numerical method of your choice.

10.58

A rectangular channel (b 5 m, S0 0.001) conveys flow at a rate of Q 3 m3/s. At a given location the depth is y1 2 m and downstream of that location the depth is y2 1.8 m. The Manning coefficient is n 0.02.

10.60

Gradually varied flow occurs over a reach of rectangular channel with n 0.013, S0 0.005, and b 2.5 m. At location 1 the depth is 1.05 m and at location 2 the depth is 1.2 m. If the two locations are 50 m apart, find the discharge and identify the profile type.

10.61

Over a portion of a very long rectangular channel of width b (Fig. P10.61), uniform flow occurs with discharge Q and depth y0. At the end of the channel the flow is terminated by a free outfall. In addition, at the outlet the channel width is reduced to b1 m for a very short distance. Determine and describe as completely as possible the changes that take place in the water surface between locations A and B. Use an E–y diagram for illustration.

(a)

Determine the distance between locations 1 and 2. (b) What type of water surface profile exists in the reach? 10.59

A rectangular channel (Fig. P10.59) has the following properties: b 1 m, S0 0.0002, n 0.017. A 120° V-notch weir is placed in the channel to measure the discharge, with the bottom notch of the weir located 0.5 m above the channel floor. Upstream of the weir the depth Y is measured to be 0.37 m. (a)

Q 5.5 m3/s, y0 0.5 m, b 3 m, b1 1.5 m. (b) Q 200 ft3/sec, y0 1.65 ft, b 10 ft, b1 5 ft.

(a)

b yo

What is the discharge in the channel?

(b) If uniform flow conditions exist very far upstream of the weir, classify the nature of the water surface.

A B

Fig. P10.61

Y

0.5 m

b=1m

Fig. P10.59

b1

Q

10.62

A sluice gate is placed in a long rectangular channel, b 4 m, n 0.014, and S0 0.0008. The depth upstream of the gate is y1 1.85 m and downstream of the gate the depth is y2 0.35 m. Neglecting losses, identify and compute the water surface profiles on either side of the gate.


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10.63

With the following data,7 identify the profiles and plot the water surface and energy grade line to scale.

Q = 3.5 b= 5

S0 = 0.001 yu = 0.80

n = 0.011 m = 2.5

L = 200 yd = 2.00

g = 9.81 c1 = 1.00

yc = y0 =

Depth 1.356 1.442

Residual 3.44E-07 4.23E-07

Station 1 2 3 4 5 6 7 8

y 0.800 0.850 0.900 0.950 1.000 1.050 1.100 1.150

A 5.600 6.056 6.525 7.006 7.500 8.006 8.525 9.056

V 6.250 5.779 5.364 4.996 4.667 4.372 4.106 3.865

E 2.791 2.552 2.366 2.222 2.110 2.024 1.959 1.911

ym

S(ym)

¢x

x

ycj

0.825 0.875 0.925 0.975 1.025 1.075 1.125

8.311E-03 6.689E-03 5.443E-03 4.472E-03 3.706E-03 3.097E-03 2.606E-03

33 33 33 32 32 31 30

33 65 98 130 162 193 223

2.044 1.961 1.883 1.809 1.740 1.674 1.611

7 8 9 10 11 12 13 14

2.000 1.975 1.950 1.925 1.900 1.875 1.850 1.825

20.000 19.627 19.256 18.889 18.525 18.164 17.806 17.452

1.750 1.783 1.818 1.853 1.889 1.927 1.966 2.006

2.156 2.137 2.118 2.100 2.082 2.064 2.047 2.030

1.988 1.963 1.938 1.913 1.888 1.863 1.838

2.770E-04 2.916E-04 3.073E-04 3.240E-04 3.417E-04 3.607E-04 3.810E-04

26 26 27 27 27 27 27

200 174 147 121 94 67 40 13

Residual 5.98E-07 3.09E-07 8.24E-07 7.46E-07 9.44E-08 2.14E-07 4.94E-07

7

The data was generated by Excel spreadsheet.


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Problems 539 10.64

With the following data,8 identify the profiles and plot the water surface and energy grade line to scale.

(a)

Q = 3.5 b = 2.5

n = 0.013 m= 0

S0 = 0.005 yu = 0.30

L = 100 yd = 0.95

g = 9.81 c1 = 1.00

yc = y0 =

Depth 0.585 0.508

Residual 8.58E-07 1.42E-07

Station 1 2 3 4 5 6 7 8 9

y 0.300 0.325 0.350 0.375 0.400 0.425 0.450 0.475 0.500

A 0.750 0.813 0.875 0.938 1.000 1.063 1.125 1.188 1.250

V 4.667 4.308 4.000 3.733 3.500 3.294 3.111 2.947 2.800

E 1.410 1.271 1.165 1.085 1.024 0.978 0.943 0.918 0.900

ym

S(ym)

¢x

x

ycj

0.313 0.338 0.363 0.388 0.413 0.438 0.463 0.488

2.154E-02 1.702E-02 1.369E-02 1.119E-02 9.272E-03 7.774E-03 6.587E-03 5.635E-03

8 9 9 10 11 13 16 29

8 17 26 36 47 60 76 104

0.985 0.932 0.884 0.840 0.799 0.762 0.727 0.694

10 11 12 13 14 15 16

0.950 0.850 0.800 0.750 0.700 0.650 0.600

7.006 6.056 5.600 5.156 4.725 4.306 3.900

0.500 0.578 0.625 0.679 0.741 0.813 0.897

0.963 0.867 0.820 0.773 0.728 0.684 0.641

0.900 0.825 0.775 0.725 0.675 0.625

9.698E-04 1.236E-03 1.474E-03 1.781E-03 2.183E-03 2.725E-03

24 13 13 14 16 19

100 76 64 51 36 21 2

(b)

Q = 125 b= 8

n = 0.013 m= 0

S0 = 0.005 yu = 1.00

L = 300 yd = 3.00

Residual 6.45E-07 4.60E-07 6.34E-07 3.55E-07 2.00E-07 4.40E-07 9.05E-07 2.22E-07

g = 32.2 c1 = 1.49

yc = y0 =

Depth 1.965 1.710

Residual 1.04E-07 4.30E-04

Station 1 2 3 4 5 6 7 8

y 1.000 1.100 1.200 1.300 1.400 1.500 1.600 1.700

A 8.000 8.800 9.600 10.400 11.200 12.000 12.800 13.600

V 15.625 14.205 13.021 12.019 11.161 10.417 9.766 9.191

E 4.791 4.233 3.833 3.543 3.334 3.185 3.081 3.012

ym

S(ym)

¢x

x

ycj

1.050 1.150 1.250 1.350 1.450 1.550 1.650

2.155E-02 1.634E-02 1.269E-02 1.007E-02 8.135E-03 6.674E-03 5.549E-03

34 35 38 41 48 62 126

34 69 107 148 195 258 384

3.311 3.102 2.914 2.744 2.589 2.447 2.317

9 10 11 12 13 14 15

3.000 2.800 2.600 2.400 2.200 2.000 1.965

37.500 33.600 29.900 26.400 23.100 20.000 19.478

3.333 3.720 4.181 4.735 5.411 6.250 6.417

3.173 3.015 2.871 2.748 2.655 2.607 2.605

2.900 2.700 2.500 2.300 2.100 1.983

1.105E-03 1.349E-03 1.674E-03 2.121E-03 2.751E-03 3.247E-03

40 39 37 32 21 1

300 260 220 183 151 129 128

Residual 3.50E-07 8.20E-07 5.79E-07 3.35E-07 2.31E-07 3.00E-07 2.58E-07

8

The data was generated by Excel spreadsheet.


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Chapter 10 / Flow in Open Channels (a)

Identify the type of profile that exists in the reservoir. (b) Compute the distance between A and B. (c) Sketch the water surface and energy grade line.

Consider a backwater curve situated upstream of a dam in a rectangular reservoir (Fig. P10.65). Given data are: normal depth y0 3.92 m, Q 125 m3/s, S0 0.0004, n 0.025, yA 11 m, yB 12 m

A Q B

Fig. P10.65

10.66

A rectangular channel has a change in slope as shown in Fig. P10.66. The channel is 3.66 m wide, with n 0.017, S02 0.00228, and Q 15.38 m3/s.

(a)

In the three channels shown in Fig. P10.67, normal depths and critical depths are indicated by dashed (- - -) and dotted (. . . ) lines, respectively. Sketch one possible composite profile for each system, and identify all gradually varied surfaces.

Determine the depth that must exist in the downstream channel for a hydraulic jump to terminate at uniform flow conditions.

(b) If y01 0.6 m, calculate the length to the jump, Lj, using several increments of depth in a step calculation.

(c)

10.67

(a)

Sketch the water surface and energy grade line, and identify all gradually varied flow profiles. (b) y 01 y 02

Lj

Fig. P10.66

(c)

Fig. P10.67


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Problems 541

10.68

hydraulic jump that may occur downstream of the dam off the river bed for the given design conditions to prevent erosion of the river bed. The depth behind the jump at the end of the apron is y2 0.75 m. Determine the following: (a) Power in kilowatts dissipated by the jump (b) Design length of the apron (c) The gradually-varied-flow profiles upstream of the dam, and between the toe and the hydraulic jump

A design flow of Q 19 m3/s occurs in a river that can be approximated as a wide rectangular channel with the following properties: b 20 m, S0 0.0005, n 0.016. A small dam is to be placed in the channel as part of a flood control plan (Fig. P10.68). The height of the dam is h 6 m, and the depth of flow at the toe of the dam is ytoe 0.10 m. A concrete spillway apron (n 0.014) is to be situated downstream of the dam. The purpose of the apron is to keep any

Q

y2

y toe

h

L

Fig. P10.68

10.69 A very wide rectangular channel has the following properties: S0 0.0005, n 0.017, q 1.2 m2/s. At some location, say location A, the depth is 0.65 m, and at location B, the depth is 0.90 m. (a) Is location B upstream or downstream of location A? (b) Using the varied flow function determine the distance between locations A and B. 10.70 An estuary to a river empties to the ocean. The estuary can be approximated as a very wide channel with a depth of 7 m at its outlet, and with S0 0.0001, n 0.015. If the discharge into the

ocean from the estuary is q 1.5 m3/s per meter width, determine the distances upstream from the ocean where the depth is equal to 6, 5, 4, and 2 m. 10.71 A river cross section can be approximated by two rectangles as shown in Fig. P10.71. The channel properties and dimensions are: b1 150 m, z1 4 m, n1 0.03, b2 5 m, z2 0 m, and n2 0.02. If the slope of the energy grade line is S 0.0005 and the depth y 5 m, find: (a) The energy coefficient for the composite section (b) The discharge

z1

y z2 b1

z

b2

Fig. P10.71 10.72

Consider the same river cross section of Problem 10.71. In the accompanying table, data are provided for two locations x along the river: At the downstream location x 400 m, the depth y 3.0 m when the discharge Q 280 m3/s. Using a trial-and-error procedure, determine the depth y at the upstream location x 0.

x (m) 0 400

b1 (m)

b2 (m)

z1 (m)

z2 (m)

n1

n2

115 149

107 91

16.7 17.7

15.0 15.1

0.05 0.05

0.03 0.03


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In a complex of industrial piping such as this one, engineering design requires that steady flows be analyzed for correct pipe sizing and placement of pumps. In addition, analysis is occasionally undertaken to mitigate unsteady, or transient, excitations to the system. (Marafona/Shutterstock)


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11 Flows in Piping Systems Outline 11.1 Introduction 11.2 Losses in Piping Systems 11.2.1 Frictional Losses in Pipe Elements 11.3 Simple Pipe Systems 11.3.1 Series Piping 11.3.2 Parallel Piping 11.3.3 Branch Piping 11.4 Analysis of Pipe Networks 11.4.1 Generalized Network Equations 11.4.2 Linearization of System Energy Equations 11.4.3 Hardy Cross Method 11.4.4 Network Analysis Using Generalized Computer Software 11.5 Unsteady Flow in Pipelines 11.5.1 Incompressible Flow in an Inelastic Pipe 11.5.2 Compressible Flow in an Elastic Pipe 11.6 Summary

Chapter Objectives The objectives of this chapter are to: ▲ Compare empirical frictional loss equations in piping ▲ Introduce the concept of hydraulic grade line as a variable for piping analysis ▲ Describe ad hoc methods of computing discharges and pressures in simple pipe systems ▲ Present linearized solutions (i.e., Hardy Cross method) for pipe networks using spreadsheet analysis ▲ Show use of computational software for pipe network analysis ▲ Introduce simplified analysis for unsteady flows in simple pipe systems

543


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Chapter 11 / Flows in Piping Systems

11.1 INTRODUCTION

Elements: Reaches of constant-diameter piping. Components: Valves, tees, bends, reducers, or any other device that creates a loss to the system.

Internal flows in pipelines and ducts are commonly encountered in all parts of our industrialized society. From delivering potable water to transporting chemicals and other industrial liquids, engineers have designed and constructed untold kilometers of relatively large-scale piping systems. Smaller piping units are also in abundance: in hydraulic controls, in heating and air conditioning systems, and in cardiovascular and pulmonary flow systems, to name only a few. These flows can be either steady or unsteady, uniform or nonuniform. The fluid can be either incompressible or compressible, and the piping material can be elastic, inelastic, or perhaps, viscoelastic. In this chapter we concentrate primarily on incompressible, steady flows in rigid piping. The piping system may be relatively simple, such that the variables can be solved rather easily using a calculator, or it may be sufficiently complicated that it is more convenient to use computational software. Piping systems are considered to be composed of elements and components. Basically, pipe elements are reaches of constant-diameter piping, and the components consist of valves, tees, bends, reducers, or any other device that may create a loss to the system. In addition to components and elements, pumps add energy to the system and turbines extract energy. The elements and components are linked at junctions. Fig. 11.1 illustrates several types of piping systems. Following a discussion of piping losses, several pipe systems, including series, branch, and parallel configurations, are analyzed. Attention is then directed to comprehensive network systems, where several methods of solution are presented. Most of the piping problems analyzed are those where the discharge is the unknown variable; this type of problem is classified as category 2 in Section 7.6.3. Finally, a brief introduction to unsteady flow in pipelines is presented. This topic is one that has been important for many years. It is taking on increasing importance as the demand for the construction of more cost-effective piping continues and the manner in which piping, valves, pumps, and other components interact and become more sophisticated. We address only the fundamental aspects of unsteady flows, focusing on two assumptions in a single pipe of constant diameter: an incompressible flow in an inelastic pipe, and a compressible liquid flow in an elastic pipe.

11.2 LOSSES IN PIPING SYSTEMS Losses can be divided into two categories: (a) those due to wall shear in pipe elements and (b) those due to piping components. The former are distributed along the length of pipe elements. The latter are treated as discrete discontinuities in the hydraulic and energy grade lines and are commonly referred to as minor losses; they are due primarily to separated or secondary flows. The fundamental mechanics of wall shear and development of the empirical relations relating to pipe losses are treated in Chapter 7. Minor losses are covered in detail in Section 7.6.4 and are not discussed further here. The following material focuses on the treatment of losses in the analysis of piping systems.


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Sec. 11.2 / Losses in Piping Systems 545

Flow P (a)

P

Flow demand Qe (b)

Qe

P

(c)

Fig. 11.1

Pipe systems: (a) single pipe; (b) distribution network; (c) tree network.

11.2.1 Frictional Losses in Pipe Elements It is convenient to express the pipe element frictional loss in the exponential form hL RQb

(11.2.1)


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in which hL is the head loss over length L of pipe, R is the resistance coefficient, Q is the discharge in the pipe, and b is an exponent. Depending on the formulation chosen, the resistance coefficient may be a function of pipe roughness, Reynolds number, or length and diameter of the pipe element. In particular, the Darcy–Weisbach relation, Eq. 7.6.23, can be substituted into Eq. 11.2.1. Then b 2, and the resulting expression for R is fL R 2 2gDA 8fL 25 gp D

KEY CONCEPT Frictional losses in piping are commonly evaluated using the Darcy–Weisbach or Hazen–Williams equation. The Darcy–Weisbach formulation provides a more accurate estimation.

(11.2.2)

where f is the friction factor. The characteristics of f for commercial pipe flows are developed in Section 7.6.3. Specifically, the Moody diagram, Fig. 7.13, presents a comprehensive picture of just how the friction factor varies over a wide range of Reynolds numbers and for a range of relative roughnesses. For pipe network analysis, it is convenient to express the behavior of f using approximate, equivalent empirical formulas in which the friction factor can be obtained directly in terms of the Reynolds number and relative roughness. A number of relations have been developed and shown to be reasonably accurate for engineering calculations (Benedict, 1980). In particular, the formulas of Swamee and Jain (1976) are presented in Section 7.6.3 and are shown to accurately represent the Colebrook relation, Eq. 7.6.28. The friction factor formula developed by Swamee and Jain is

冦冤

冢冣

0.9

冢冣冥冧

e 1 f 1.325 ln 0.27 5.74 D Re

2

(11.2.3)

Combining Eqs. 11.2.2 and 11.2.3, one finds that

冢

冣冦冤

冢冣

0.9

冢冣冥冧

L e 1 R 1.07 5 ln 0.27 5.74 gD D Re

2

(11.2.4)

Equations 11.2.3 and 11.2.4 are valid over the ranges 0.01 e/D 108, and 108 Re 5000. The fully rough regime, where Re has a negligible effect on f, begins at a Reynolds number given by 200D Re e兹苶f

(11.2.5)

For values of Re greater than this, the friction factor is a function only of e/D, and is given by


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冦冤

2

冢冣冥冧

e f 1.325 ln 0.27 D

(11.2.6)

Two additional expressions for pipe frictional losses, which have found wide use, are the Hazen–Williams and Chezy–Manning formulas. For water flow, the value of R in Eq. 11.2.1 for the Hazen–Williams relation is K1L R CbDm

K1 e

10.59 4.72

SI units English units

(11.2.7)

in which the exponents are b 1.85 and m 4.87, and C is the Hazen–Williams coefficient dependent only on the roughness. Values of the Hazen–Williams roughness coefficient C are given in Table 11.1. The Chezy–Manning equation is more commonly associated with openchannel flow. However, in sewage and drainage systems in particular, it has been applied to conduits flowing under surcharge, that is, under pressurized conditions. The Chezy–Manning equation was introduced in Section 7.7. For a circular pipe flowing full, Eq. 7.7.6 can be substituted into Eq. 11.2.1 and solved for R: 10.29n2L R K2D5.33

K2 e

1 2.22

SI units English units

(11.2.8)

in which n is the Manning roughness coefficient. In Eq. 11.2.1, the exponent b 2. An advantage to using Eq. 11.2.7 or Eq. 11.2.8 as opposed to Eq. 11.2.4 is that in the first two, C and n are dependent on roughness only, while in the last, f depends on the Reynolds number as well as the relative roughness. However, Eq. 11.2.4 is recommended since it provides a more precise representation of pipe frictional losses. Note that the Hazen–Williams and Chezy–Manning relations are dimensionally inhomogeneous,1 whereas the Swamee–Jain equation is dimensionally homogeneous and contains the two parameters e/D and Re, which appropriately influence the losses. Table 11.1 Nominal Values of the Hazen–Williams Coefficient C Type of pipe Extremely smooth; asbestos-cement New or smooth cast iron; concrete Wood stave; newly welded steel Average cast iron; newly riveted steel; vitrified clay Cast iron or riveted steel after some years of use Deteriorated old pipes

C 140 130 120 110 95–100 60–80

1

In a dimensionally inhomogeneous equation, the constants in the equation have units assigned to them. In a dimensionally homogeneous equation the constants are dimensionless.


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The limitations of the Hazen–Williams and Chezy–Manning formulas are demonstrated as follows. Beginning with Eq. 11.2.1, the head loss hL based on the Darcy–Weisbach resistance coefficient, Eq. 11.2.2, with the exponent b 2 can be equated with hL based on the Hazen–Williams coefficient, Eq. 11.2.7, with b 1.85. Introducing the Reynolds number to eliminate Q and solving for the friction factor f results in 1.28gK1 f C1.85D0.02(Re n)0.15

(11.2.9)

With SI units and for water at 20°C, Eq. 11.2.9 reduces to 1056 f C1.85D0.02Re0.15

(11.2.10)

Note that f is weakly dependent upon D in this equation. In a similar fashion, Eq. 11.2.8 can be substituted into Eq. 11.2.1 with b 2 and equated to hL based on the Darcy–Weisbach formulation to yield 124.5n2 f 0 D .33

(11.2.11)

Comparisons between the original Colebrook formula (Eq. 7.6.28), the formula of Swamee and Jain (Eq. 11.2.3), and the equivalent expressions for the Hazen–Williams and Manning coefficients (Eqs. 11.2.10 and 11.2.11) are shown in Fig. 11.2 for a 1-m-diameter concrete pipe flowing full with water (C 130,

Friction factor f

0.050

0.010

0.005 10 3

Colebrook Chezy–Manning Hazen–Williams Swamee–Jain 10 4

10 5

10 6

10 7

10 8

Reynolds number Re

Fig. 11.2 Comparison of several approximate formulas with the original Colebrook formula.


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n 0.012, e/D 0.00015). It is evident that the Hazen–Williams and Chezy–Manning relations are valid over a limited range of Re, and that Eq. 11.2.3 provides a more versatile and accurate estimate of pipe losses. In the remainder of this chapter most of the frictional losses will be treated using the Darcy–Weisbach formulation, Eqs. 11.2.2 to 11.2.6. Any exceptions will be noted.

Example 11.1 A pipeline is conveying 0.05 m3/s of water, at 30°C. The length of the line is 300 m, and the diameter is 0.25 m. Estimate the head loss due to friction using the three formulas (a) Darcy–Weisbach (e 0.5 mm), (b) Hazen–Williams (C 110), and (c) Chezy–Manning (n 0.012). Solution The kinematic viscosity of water at 30°C is n 0.804 106 m2/s. The relative roughness, velocity, and Reynolds number are computed to be e 0.002 D

V 1.02 m앛s

Re 3.17 105

(a) Substitute values into Eq. 11.2.4: 300 R 1.07 5 9.81 (0.25) {ln[0.27 0.002 5.74 (3.17 105)0.9]}2 610 Then, with Eq. 11.2.1, the friction loss based on the Darcy–Weisbach formulation is hL RQ2 610 (0.05)2 1.52 m (b) For the Hazen–Williams formulation, use Eq. 11.2.7: 10.59 300 454 R (110)1.85(0.25)4.87 hL 454 (0.05)1.85 1.78 m (c) Use Eq. 11.2.8 for the Chezy–Manning formulation: 10.29(0.012)2(300) R 719 1 (0.25)5.33 hL 719 (0.05)2 1.80 m Part (a) with hL 1.52 m is the most accurate. The Hazen–Williams result is 17% too high, and the Chezy–Manning result is 18% too high.


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11.3 SIMPLE PIPE SYSTEMS The analysis of a single pipeline is presented in Section 7.6.4; it would be appropriate here for the reader to review the three categories of pipe problems in that section. For more complex piping systems, the methodology is similar. With relatively simple networks, such as series, parallel, and branching systems, ad hoc solutions can be developed that are suitable for use with calculators, spreadsheet algorithms, or computational software. Such approaches are relevant since they make use of the solver’s ingenuity and require an understanding of the nature of the flow and piezometric head distributions for the particular piping arrangement. For systems of greater complexity, an alternative means of solving such problems is the Hardy Cross method, presented in Section 11.4. The fundamental principle in the ad hoc approach is to identify all of the unknowns and write an equivalent number of independent equations to be solved. Subsequently, the system is simplified by eliminating as many unknowns as possible and reducing the problem to a series of single pipe problems—either category 1 or category 2, Section 7.6.3.

11.3.1 Series Piping Consider the series system shown in Fig. 11.3. It consists of N pipe elements with a specified number of minor-loss components K associated with each ith pipe element. A single minor loss is described in Section 7.6.5 to be equal to hL KV 2/2g. It is convenient here to express the minor loss in terms of the discharge rather than the velocity, so that hL KQ2/2gA2. For many flow situations, it is common practice to neglect the kinetic-energy terms at the inlet and outlet; they would be significant only if the velocities were relatively high. Assuming that the exponent in Eq. 11.2.1 is b 2, the energy equation applied from location A to location B of Fig. 11.3 is p

冢g z冣

KEY CONCEPT The

A

principles of continuity and energy are used to analyze pipe systems. The predicted parameters are discharge and piezometric head.

p z g

冢

冣

B

K K R1 2 Q 12 R2 2 Q 22 2gA1 2gA 2

冢

冣

冢

冣

K

QN2 冢RN 2gA 2 冣 N

K 兺 Ri 2 Q 2i i1 2gAi N

冢

冣

(11.3.1)

in which Ri is the resistance coefficient for pipe i.

[1]

[i ]

[2]

A

[N ]

B Qi

Fig. 11.3

Series piping system.


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Sec. 11.3 / Simple Pipe Systems 551

The statement of continuity for the series system is that the discharge in every element is identical, or Q1 Q2

Qi QN Q

(11.3.2)

Replacing Qi with Q, Eq. 11.3.1 becomes p

p z A g

冢g z冣

冢

冣

B

K

N

Q 冤兺冢R 2gA 冣冥 i

i1

2

(11.3.3)

2 i

Equation 11.2.4 can be substituted for Ri, or alternatively, one may use Eq. 11.2.3 or the Moody diagram, Fig. 7.13, to obtain values of the friction factor. The reader should recognize that in a series system, the discharge remains constant from one pipe element to another, and the losses are accumulative; that is, they are the sum of the minor loss components and the pipe frictional losses. For a category 1 problem, the right-hand side of Eq. 11.3.3 is known and the solution is straightforward. For a category 2 problem, in which Q is unknown, a trial-and-error solution is required, since the Reynolds number, in terms of the unknown discharge (Re 4Q/p D), is present in the friction factor relation. Note that if flow in the fully rough zone is assumed, f is independent of Q and Eq. 11.3.3 reduces to a quadratic in Q. A category 3 problem is not encountered in this type of analysis. A series piping system with minor losses and constantdiameter piping is applied to a category 1 problem in Example 7.13. The solution for a somewhat more complex series system is illustrated in Example 11.2.

Example 11.2 For the system shown in Fig. E11.2, find the required power to pump 100 L/s of liquid (S 0.85, n 105 m2/s). The pump is operating at an efficiency h 0.75. Pertinent data are given in the figure. Line 1: L 10 m, D 0.20 m, e 0.05 mm, K1 0.5, Ky 2 Line 2: L 500 m, D 0.25 m, e 0.05 mm, Ke 0.25, K2 1

el 20 m Ke

B

K2 Line 2

A el 10 m K1

Line 1 Ke

P Kυ

Fig. E11.2

(continued)


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Solution This is a category 1 problem. The energy relation, Eq. 11.3.3, for the system is p

冢g z冣

A

p HP z g

冢

冣

B

K1 Ky 2Ke K2 2 R1 R2 Q 2gA 12 2gA 22

冤

冥

The resistance coefficients, R1 and R2, are calculated with Eq. 11.2.4 after first evaluating Re and e/D: 4Q 4 0.10 Re1 6.37 104 pD1n p 0.20 105

冢D冣

1

0.05 0.00025 200

4Q 4 0.10 Re2 5.09 104 pD2n p 0.25 105

冢D冣

2

0.05 0.0002 250

冢

10 R1 1.07 9.81 0.205

e

e

冣

{ln[0.27 0.00025 5.74 (6.37 104)0.9]}2 53.4

冢

500 R2 1.07 5 9.81 0.25

冣

{ln[0.27 0.0002 5.74 (5.09 104)0.9]}2 904 The minor loss coefficient terms are calculated to be 0.5 2 K1 Ky 129.1 2gA 21 2 9.81 3p/4 0.202 4 2 2 0.25 1 2Ke K2 31.7 2gA22 2 9.81 3p/4 0.252 4 2 Substitute these values into the energy equation and obtain 200 103 0 10 HP 20 (53.4 129.1 904 31.7) 0.12 0.85 9800 This relation reduces to 10 HP 24 20 11.2 Solving for the head across the pump gives HP 45.2 m. The required input power is gQHP ˙ W P h

(9800 0.85) 0.10 45.2 0.75

5.0 104 W

or

50 kW


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11.3.2 Parallel Piping A parallel piping arrangement is shown in Fig. 11.4; it is essentially an arrangement of N pipe elements joined at A and B with K minor loss components associated with each pipe element i. The continuity equation applied at either location A or B is given by N

Q 兺 Qi

(11.3.4)

i1

The algebraic sum of the energy grade line around any defined loop must be zero. As in the case of series piping, it is customary to assume that V 2/2g (p/g z). Hence, for any pipe element i, the energy equation from location A to B is p

冢g z冣

A

p z g

冢

冣

B

K Ri 2 Q 2i 2gAi

冢

冣

i 1, . . . , N

(11.3.5)

The unknowns in Eqs. 11.3.4 and 11.3.5 are the discharges Qi and the difference in piezometric head between A and B; the discharge Q into the system is assumed known. It is possible to convert the minor loss terms using an equivalent length as defined in Section 7.6.4. For each pipe element i the equivalent length Le for K minor loss components is Di (Le)i K fi

(11.3.6)

Thus Eq. 11.3.5 simplifies to the form

冢g z冣 P

冢

冣

P z A g

B

苶 Ri Q 2i

(11.3.7)

[1] [2]

A

B

Q [i]

[N ]

Fig. 11.4

Parallel piping system.


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in which the modified pipe resistance coefficient 苶 Ri is given by 8fi[Li (Le)i] R 苶i 2 gp D 5i

(11.3.8)

Note that the right-hand side of Eq. 11.3.8 is equivalent to the term Ri K/(2gA2i ), where Eq.11.2.2 expresses Ri . A solution employing the method of successive substitution is developed in the following manner. Define the variable W to be the change in hydraulic grade line between A and B; that is, W (p/g z)A (p/g z)B. Then Eq. 11.3.7 can be solved for Qi in terms of W as Qi

冪莦W苶R

(11.3.9)

i

Equations 11.3.4 and 11.3.9 are combined to eliminate the unknown discharges Qi, resulting in N

Q兺

i1

冪莦

N W 1 兹W 苶兺 i1 兹苶 Ri 苶 R 苶i

(11.3.10)

The remaining unknown W is taken out of the summation sign since it is the same in all pipes. Solving for W in Eq. 11.3.10 results in

W

冢兺冣 Q

N

i1

1 兹苶 R 苶i

2

(11.3.11)

An iterative procedure can be formulated to solve for W and the discharges Qi as follows: 1. Assume flows in each line to be in the completely rough zone, and compute an initial estimate of the friction factors in each line using Eq. 11.2.6. 2. Compute R 苶i for each pipe and evaluate W with Eq. 11.3.11. 3. Compute Qi in each pipe with Eq. 11.3.9. 4. Update the estimates of the friction factors in each line using the current values of Qi and Eq. 11.2.3. 5. Repeat steps 2 to 4 until the unknowns W and Qi do not vary within a desired tolerance. Note that if friction factors are in the completely rough zone so that they are independent of the discharge and therefore constant, steps 4 and 5 are unnecessary and a solution results on the first iteration. The technique is illustrated in Example 11.3.


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Example 11.3 Find the distribution of flow and the drop in hydraulic grade line for the three-parallelpipe arrangement shown in Fig. E11.3. Use variable friction factors with n 106 m2/s. The total water discharge is Q 0.020 m3/s.

Q

[1]

Pipe

L (m)

D (m)

e (mm)

K

[2]

1 2 3

100 150 200

0.05 0.075 0.085

0.1 0.2 0.1

10 3 2

[3]

Fig. E11.3

Solution The initial estimates of f are based on Eq. 11.2.6. Preliminary calculations provide the following:

Pipe

e/D

f (Eq. 11.2.6)

Le (Eq. 11.3.6)

R 苶 (Eq. 11.3.8)

1 2 3

0.002 0.0027 0.0012

0.023 0.025 0.021

21.7 9.0 8.1

7.40 105 1.38 105 8.14 104

Apply Eq. 11.3.11, and the first estimate of W is

冤

0.020 W (7.40 105)1앛2 (1.38 105)1앛2 (8.14 104)1앛2

冥

2

7.39 m Then with Eq. 11.3.9, estimates of Qi are made:

冢

冣

0.00316 m3앛s

冢

冣

0.00732 m3앛s

冢

冣

0.00953 m3앛s

7.39 Q1 5 7.40 10 7.39 Q2 5 1.38 10 7.39 Q3 4 8.14 10

1앛2

1앛2

1앛2

A continuity check is made using Eq. 11.3.4: 3

兺 Qi 0.00316 0.00732 0.00953 0.0200 m3앛s

i1

Even though the sum satisfies the solution, another iteration will be carried out to study the convergence of the solution technique. First, the R 苶-values are updated using Eq. 11.2.3 to evaluate the friction factors. (continued)


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Pipe

4Q Re pDn

f (Eq. 11.2.3)

R 苶 (Eq. 11.3.8)

1 2 3

8.05 104 1.24 105 1.43 105

0.026 0.027 0.022

8.20 105 1.49 105 8.51 104

Evaluating W using Eq. 11.3.11 yields W 7.88 m, and application of Eq. 11.3.9 gives the new estimates of Qi: Q1 0.00310 m3앛s

Q2 0.00727 m3앛s

Q3 0.00962 m3앛s

A continuity check shows that 3

兺 Qi 0.0200 m3앛s

i1

which is the same as in the first iteration.

A solution of Example 11.3 using Mathcad is shown in Appendix F, Fig. F1.

11.3.3 Branch Piping The branching network, illustrated in Fig. 11.5a, is made up of three elements connected to a single junction. In contrast to the parallel system shown in Fig. 11.4, no closed loops exist. In the analysis, one assumes the direction of flow in each element; then the energy equation for each element is written using an equivalent length to account for minor losses:

冢g z冣 p

A

冢

冣

苶 R1Q 21

(11.3.12)

冢

冣

苶 R2Q 22

(11.3.13)

冢

冣

苶 R3Q 32

(11.3.14)

p z g

冢g z冣

B

p z g

冢g z冣

B

p z g

p

p

B

C

D

The piezometric heads at locations A, C, and D are considered known. The unknowns are the piezometric head at B and the discharges Q1, Q2, and Q3. The additional relation is the continuity balance at location B, which is Q1 Q2 Q3 0

(11.3.15)

Thus there are four equations with four unknowns. One convenient ad hoc method of solution is outlined below and illustrated in Example 11.4: 1. Assume a discharge Q1 in element 1 (with or without a pump). Establish the piezometric head H at the junction by solving Eq. 11.3.12 (a category 1 problem). Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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A

D [1] [2] Q2

Q1 B

Q3 [3] C (a)

C

D Q2

Q3

[2]

Q1 A

[3]

P B [1] (b)

Fig. 11.5

Branch piping systems: (a) gravity flow; (b) pump-driven flow.

2. Compute the discharge Qi in the remaining branches using Eqs. 11.3.13 and 11.3.14 (a category 2 problem). 3. Substitute the Qi into Eq. 11.3.15 to check for continuity balance. Generally, the flow imbalance at the junction Q will be nonzero. In Eq. 11.3.15, Q Q1 Q2 Q3. 4. Adjust the flow Q1 in element 1 and repeat steps 2 and 3 until Q is within desired limits. If a pump exists in pipe 1 (see Fig. 11.5b), Eq. 11.3.12 is altered in the manner

冢g z冣 p

A

冢

冣

p z g

B

HP 苶 R1Q 21

(11.3.16)

An additional unknown, namely, the pump head HP, is introduced. The additional necessary relationship is the head-discharge curve for the pump; see, for example, Fig. 7.18. The solution can proceed in a manner similar to that described in Example 11.5. It may be convenient to follow the solution graphically by plotting the assumed discharge in line 1 versus either the piezometric head at B or the imbalance of flow at B. Such a step is helpful to determine in what manner the discharge in line 1 should be altered for the next iteration. An alternative method of solution for a single branching system is to eliminate all of the variables except the piezometric head H ( p/g z) at the junction, location B in Fig. 11.5. Then a numerical solving technique can be employed. An additional requirement is to assume the direction of flow in each pipe. It may Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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become necessary to correct the sign in one or more of the equations if during the solution, H moves from below one of the reservoir elevations to above it, or vice versa. Example 11.4 illustrates the procedure. Examples 11.4 and 11.5 represent a level of complexity that one may not want to exceed employing a calculator-based solution. To include additional pumps, reservoirs or pipes would make either ad hoc or simplified numerical approaches too cumbersome. For such systems, the more generalized network analysis described in Section 11.4 is recommended.

Example 11.4 For the three-branch piping system shown in Fig. E11.4 we have the following data: el 20 m el 13 m

Pipe

L (m)

D (m)

f

K

1 2 3

500 750 1000

0.10 0.15 0.13

0.025 0.020 0.018

3 2 7

[2]

el 5 m

[3] [1]

Fig. E11.4 Determine the flow rates Qi and the piezometric head H at the junction. Assume constant friction factors. Solution The equivalent lengths and resistance coefficients are 0.10 (Le)1 3 12 m 0.025

8 0.025 512 R1 1.06 105 s2앛m5 苶 9.81 p2 (0.10)5

0.15 (Le)2 2 15 m 0.020

8 0.020 765 R2 1.66 104 s2앛m5 苶 9.81 p2 (0.15)5

0.13 (Le)3 7 51 m 0.018

8 0.018 1051 R3 4.21 104 s2앛m5 苶 9.81 p2 (0.13)5

With the flow directions assumed as shown, the energy equation is written for each pipe and solved for the unknown discharge: H5 Q1 R 苶1

冢

20 H Q2 R 苶2

冣

冢

1앛2

冣

1앛2

H 13 Q3 R 苶3

冢

冣

1앛2

The continuity equation is Q1 Q2 Q3 0. Eliminating Q1, Q2, and Q3 with the energy relations results in an algebraic equation in terms of H: H5 „(H) 5 1.06 10

冢

冣

1앛2

20 H 4 1.66 10

冢

冣

1앛2

H 13 4 4.21 10

冢

冣

1앛2

0


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Even though this can be solved as a quadratic, the method of false position is chosen to compute H, which would be required if the friction factors varied. The procedure is presented in Example 10.11. In the present example the recurrence formula is Hl „(Hu) Hu „(Hl) Hr „(Hu) „(Hl) The solution is shown in the table below. Note that with the initial guesses of Hl and Hu, the sign conventions in „ require that 20 H 13. Iteration continues until the convergence criterion shown in the last column becomes less than the arbitrary value of 0.005.

Iteration 1 2 3

Hu

„(Hu)

Hl

„(Hl)

Sign of Hrnew Hrold „(Hl) „(Hr) old Hr

冨

„(Hr)

Hr

18 13 0.01100 0.01185 15.59 0.00154 15.59 13 0.00154 0.01185 15.29 0.000384 15.29 13 0.000384 0.01185 15.22 0.000112

冨

0.019 0.0046

Hence H ⯝ 15.2 m. The discharges are now computed: 15.2 5 Q1 5 1.06 10

1앛2

20 15.2 Q2 4 1.66 10

1앛2

15.2 13 Q3 4 4.21 10

1앛2

冢

冣

冢

冣

冢

冣

0.0098 m3앛s 0.0170 m3앛s 0.0072 m3앛s

Note that continuity is satisfied.

Example 11.5 For the system shown in Fig. E11.5, determine the flow distribution Qi of water and the piezometric head H at the junction. The fluid power input by the pump is constant, equal to g QHP 20 kW. Assume constant friction factors. el 30 m el 15 m el 10 m

[2] P [1]

B

[3]

Pipe

L (m)

D (m)

f

K

1 2 3

50 100 300

0.15 0.10 0.10

0.02 0.015 0.025

2 1 1

Fig. E11.5 Solution The equivalent lengths and resistance coefficients are computed from Eqs. 11.3.6 and 11.3.8 to be (continued)


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0.15 (Le)1 2 15 m 0.02

8 0.02 65 3 2 5 R1 苶 9.81 p2 (0.15)5 1.42 10 s 앛m

0.10 (Le)2 1 6.7 m 0.015

8 0.015 106.7 R2 1.32 104 s2앛m5 苶 9.81 p2 (0.10)5

0.10 (Le)3 1 4 m 0.025

8 0.025 304 R3 6.28 104 s2앛m5 苶 9.81 p2 (0.10)5

Assume flow directions as shown. The energy equation for pipe 1 from the reservoir to the junction B is z1 HP H 苶 R1Q 21 in which H is the piezometric head at B. Substituting in known parameters and solving for H results in 20 103 H 10 1.42 103Q 21 9800Q1 2.04 10 1420Q 12 Q1 An iterative solution is shown in the accompanying table. For each iteration, a value of Q1 is estimated. Then H is calculated and Q2 and Q3 are evaluated from the relations H z2 Q2 R2 苶

冢

冣

H 30 4 1.32 10

1앛2

H z3 Q3 R3 苶

冣

H 15 4 6.28 10

1앛2

冢

1앛2

1앛2

冢

冣

冢

冣

In the last column of the table, a continuity balance is employed to check the accuracy of the estimate of Q1. The third estimate of Q1 is based on a linear interpolation by setting Q 0 and using values of Q1 and Q from the first two iterations. Iteration

Q1

H

Q2

Q3

Q Q1 Q2 Q3

1 2 3

0.050 0.055 0.054

47.25 42.80 43.64

0.0362 0.0311 0.0322

0.0227 0.0210 0.0214

0.0089 0.0029 0.0004

The approximate solution is H 43.6 m, Q1 54 L/s, Q2 32 L/s, and Q3 21 L/s. If greater precision is desired, a solution should be employed similar to that shown in Example 11.4.

Solutions for Examples 11.4 and 11.5 using Mathcad and MATLAB are provided in Appendix F, Figs. F2 to F5. In those solutions note that the term 苶 RQ2 is replaced by R 苶Q앚Q앚, which automatically accounts for changes in flow direction. In addition to the computational software solutions in Appendix F, the more general pipe network solutions described in Section 11.4 can be applied to the problems described in this section.


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Sec. 11.4 / Analysis of Pipe Networks 561

11.4 ANALYSIS OF PIPE NETWORKS Piping systems more complicated than those considered in Section 11.3 are best analyzed by formulating the solution for a network. Before considering a generalized set of network equations, it is worthwhile to examine a specific example of a piping network to observe the degree of complexity involved. Fig. 11.6a shows a relatively simple network consisting of seven pipes, two reservoirs, and one pump. The hydraulic grade lines at A and F are assumed known; these locations are termed fixed-grade nodes. Outflow demands are present at nodes C and D. Nodes C and D, along with nodes B and E are called

F QD

D

[6] [7]

[2] E [3] [1] [5]

B [4] C

P

A

QC

(a) D F

Interior loop E

I II B A

C (b)

Pseudo loop D F E B C

A

(c)

Fig. 11.6 Representative piping network: (a) assumed flow directions and numbering scheme; (b) designated interior loops; (c) path between two fixed-grade nodes.


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interior nodes or junctions. Flow directions, even though not initially known, are assumed to be in the directions shown. The system equations are given as follows: 1. Energy balance for each pipe (seven equations): HA HB HP(Q1) 苶 R1Q 21

HC HE 苶 R5Q 52

HB HD 苶 R2Q 22

HE HD 苶 R6Q 62

R3Q 32 HC HD 苶

HF HE 苶 R7Q 72

(11.4.1)

HB HC 苶 R4Q 42 2. Continuity balance for each interior node (four equations): Q1 Q2 Q4 0 Q2 Q3 Q6 QD Q4 Q3 Q5 QC

(11.4.2)

Q5 Q6 Q7 0 3. Approximation of pump curve (one equation): HP(Q1) a0 a1Q1 a2Q 21

(11.4.3)

in which a0, a1, and a2 are known constants. The unknowns are Q1, . . . , Q7, HB, HC, HD, HE, and HP. Thus there are 12 unknowns and 12 equations to solve simultaneously. Since the energy equations and the pump equation are nonlinear, it is necessary to resort to some type of approximate numerical solution. The 12 equations can be reduced in number by combining the energy equations along special paths. Let the drop in hydraulic grade line for any pipe element i be designated as Wi. Then Wi 苶 RiQ 21

(11.4.4)

For the system under consideration, two closed paths, or interior loops, can be identified (see Fig. 11.6b). Flow is considered positive in a clockwise sense around each loop. Energy balances, written around loops I and II, are W6 W3 W5 0 W3 W2 W4 0

(11.4.5)

To account for the flow in pipes 1 and 7, a path can be defined along nodes A, B, D, E, and F in Fig. 11.6c. Then with the addition of the pump head, the energy balance from A to F is


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HA HP W1 W2 W6 W7 HF

(11.4.6)

Note that the path energy equation connects two fixed-grade nodes. Such a path is sometimes termed a pseudoloop, since an imaginary pipe with infinite resistance, or no flow, can be considered to connect the two reservoirs. The imaginary pipe is shown by a dotted line in Fig. 11.6c. Substituting the pump equation and the friction equation into the energy relations above results in the following reduced set of equations: R R5Q 52 苶 R6Q 62 0 苶3Q 32 苶 R R3Q 32 苶 R4Q 42 0 苶2Q 22 苶 R R2Q 22 苶 R6Q 62 苶 R7Q 72 HA HF 0 苶1Q 21 (a0 a1Q1 a2Q 21 ) 苶 Q1 Q2 Q4 0 Q2 Q3 Q6 QD Q4 Q3 Q5 QC Q5 Q6 Q7 0 (11.4.7) There are now seven unknowns (Q1, . . . ,Q7) and seven equations to solve. The energy relations are nonlinear since the loss terms and the pump head are represented as polynomials with respect to the discharges.

11.4.1 Generalized Network Equations Networks of piping, such as those shown in Fig. 11.6, can be represented by the following equations. 1. Continuity at the jth interior node: ()jQj Qe 0

(11.4.8)

in which the subscript j refers to the pipes connected to a node, and Qe is the external demand. The algebraic plus or minus sign convention pertains to the assumed flow direction: Use the positive sign for flow into the junction, and the negative sign for flow out of the junction. 2. Energy balance around an interior loop: ()iWi 0

(11.4.9)

in which the subscript i pertains to the pipes that make up the loop. There will be a relation for each of the loops. Here it is assumed that


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there are no pumps located in the interior of the network. The plus sign is used if the flow in the element is positive in the clockwise sense; otherwise, the minus sign is employed. 3. Energy balance along a unique path or pseudoloop connecting two fixed-grade nodes: ()i[Wi (HP)i] H 0

(11.4.10)

where H is the difference in magnitude of the two fixed-grade nodes in the path ordered in a clockwise fashion across the imaginary pipe in the pseudoloop. The term (HP)i is the head across a pump that could exist in the ith pipe element. If F is the number of fixed-grade nodes, there will be (F 1) unique path equations. The plus and minus signs in Eq. 11.4.10 follow the same argument given for Eq. 11.4.9. Let P be the number of pipe elements in the network, J the number of interior nodes, and L the number of interior loops. Then the following relation will hold if the network is properly defined: PJLF1

(11.4.11)

In Fig. 11.6 of the introductory example, J 4, L 2, F 2, so that P 4 2 2 1 7. An additional necessary formulation is the relation between discharge and loss in each pipe; it is K W RQ b 2 Q 2 2gA

(11.4.12)

If the minor losses can be defined in terms of an equivalent length, Eq. 11.4.12 can be replaced by2 W苶 RQb

(11.4.13)

An approximate pump head-discharge representation is given by the polynomial HP(Q) a0 a1Q a2Q2

(11.4.14)

2

In Eq. 11.4.13, the exponent b is used, since the Hazen–Williams formula is commonly employed in pipe network analysis. The equivalent length for use in the Hazen–Williams formulation is given by K Le 0.8106 D0.87C1.85Q0.15 K1 Note that it is necessary to estimate the discharge when using this equation. The Darcy–Weisbach equation with b 2 is recommended. An alternative to using the equivalent-length concept is to account for pipe friction and minor losses separately in each line, making use of Eq. 11.4.12.


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The coefficients a0, a1, and a2 are assumed known; typically, they can be found by substituting three known data points from a specified pump curve and solving the three resulting equations simultaneously. In place of defining the pump headdischarge curve, an alternative means of including a pump in a line is to specify the useful power the pump puts into the system. The useful, or actual, power ˙ is assumed to be constant and allows H to be represented in the manner W f P ˙ W f HP(Q) gQ

(11.4.15)

This equation is particularly useful when the specific operating characteristics of a pump are unknown.

11.4.2 Linearization of System Energy Equations Equation 11.4.10 is a general relation that can be applied to any path or closed loop in a network. If it is applied to a closed loop, H is set equal to zero, and if no pump exists in the path or loop, (HP)i is equal to zero. Note that Eq. 11.4.9 can be considered to be a subset of Eq. 11.4.10. In the following development, Eq. 11.4.10 will be used to represent any loop or path in the network. Define the function f(Q) to contain the nonlinear terms W(Q) and HP(Q) in the form f(Q) W(Q) HP(Q) 苶 RQb HP(Q)

(11.4.16)

Equation 11.4.16 can be expanded in a Taylor series as df f(Q) f(Q0) dQ

d 2f (Q Q0) 2 Q0 dQ

冨

(Q Q0)2 Q0 2

冨

(11.4.17)

in which Q0 is an estimate of Q. To approximate f(Q) accurately, Q0 should be chosen so that the difference (Q Q0) is numerically small. Retaining the first two terms on the right-hand side of Eq. 11.4.17, and using Eq. 11.4.16, yields

冤

冨冥(Q Q )

dHP f(Q) ⯝ R RQ b01 苶Q b0 HP(Q0) b 苶 dQ

Q0

0

(11.4.18)

Note that the approximation to f(Q) is now linear with respect to Q. The parameter G is introduced as

冨

dHP G bR 苶Q b01 dQ

Q0

(11.4.19)

Using Eq. 11.4.14 to represent the pump head, Eq. 11.4.19 becomes


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G bR 苶Q b0 1 (a1 2a2Q0)

(11.4.20)

Alternatively, with Eq. 11.4.15 substituted into Eq. 11.4.19, one has ˙f W G bR 苶Q b0 1 2 gQ 0

(11.4.21)

Substituting Eq. 11.4.19 into Eq. 11.4.18 gives KEY CONCEPT Complex pipe systems require special techniques to solve for discharges and piezometric heads. One of them is the Hardy Cross method of solution.

f(Q) 苶 RQ b0 HP(Q0) (Q Q0)G W0 HP0 (Q Q0)G

(11.4.22)

in which W0 W(Q0) and HP0 HP(Q0). Finally, Eq. 11.4.22 is substituted into Eq. 11.4.10 to produce the linearized loop or path energy equation ()i[(W0)i (HP0)i] [Qi (Q0)i]Gi H 0

(11.4.23)

The second term does not contain the plus or minus sign since Gi is a monotonically increasing function of the flow correction. Since Q in the relations above can assume positive or negative values, Q b0 and Q b01 are often replaced by Q0앚Q0앚b1 and 앚Q0앚b1, respectively, in solution algorithms. This is done, for example, in Eq. 11.4.23, which forms the basis for the Hardy Cross solution outlined below.

11.4.3 Hardy Cross Method In Eq. 11.4.23, let (Q0)i be the estimates of discharge from the previous iteration, and let Qi be the new estimates of the discharge. Define a flow adjustment Q for each loop to be Q Qi (Q0)i

(11.4.24)

The adjustment is applied independently to all pipes in a given loop. Hence Eq. 11.4.23 can be written as ()i[(W0)i (HP0)i] QGi H 0

(11.4.25)

Solving for Q, one has ()i[(W0)i (HP0)i] H Q Gi

(11.4.26)

It is necessary for the algebraic sign of Q to be positive in the direction of normal pump operation; otherwise, the pump curve will not be represented properly and


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Eq. 11.4.26 will be invalid. Furthermore, it is important that the discharge Q through the pump remain within the limits of the data used to generate the curve. For a closed loop in which no pumps or fixed-grade nodes are present, Eq. 11.4.26 reduces simply to the form ()i(W0)i Q Gi

(11.4.27)

The Hardy Cross iterative solution is outlined in the following steps: 1.

Assume an initial estimate of the flow distribution in the network that satisfies continuity, Eq. 11.4.8. The closer the initial estimates are to the correct values, the fewer will be the iterations required for convergence. One guideline to use is that in a pipe element, as R 苶 increases, Q decreases. 2. For each loop or path, evaluate Q with Eq. 11.4.26 or 11.4.27. The numerators should approach zero as the loops or paths become balanced. 3. Update the flows in each pipe in all loops and paths, that is, from Eq. 11.4.24. Qi (Q0)i Q

(11.4.28)

The term Q is used, since a given pipe may belong to more than one loop; hence the correction will be the sum of corrections from all loops to which the pipe is common. 4. Repeat steps 2 and 3 until a desired accuracy is attained. One possible criterion to use is 앚Qi (Q0)i앚 앚Qi앚

(11.4.29)

in which is an arbitrarily small number. Typically, 0.001 0.005. The Hardy Cross method of analysis is a simplified version of the method of successive approximations applied to a set of linearized equations. It does not require the inversion of a matrix; hence it can be used to solve relatively small networks using either a calculator or a spreadsheet algorithm on a personal computer. The continuity relations, Eq. 11.4.8, are in a sense “decoupled” from the solution of the energy relations, Eq. 11.4.26. Continuity is satisfied initially with assumed flows and remains satisfied throughout the solution process. Essentially, one computes a correction Q to the flows Qi in each loop separately, and then applies Q to the entire network to bring flow through the loops into closer balance. In effect this is a superposition type of solution. Since the flow corrections Q are applied to each loop independently, convergence may not be rapid.


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Example 11.6 For the piping system shown in Fig. E11.6a, determine the flow distribution and piezometric heads at the junctions using the Hardy Cross method of solution. Assume that losses are proportional to Q2. elA = 50 m A

elB = 30 m B [1] C

R1 = 100

D

[2]

R3 = 200 R8 = 300

R2 = 500 [6]

[8] R6 = 300

[4] E R4 = 100

[5]

[7]

F

[3]

R5 = 400

G

R7 = 400

150 L/s

150 L/s (a)

A B

I

Q1 Q2

Q3 Q4

Q6

II

III

Q8

Q5 Q7 (b)

A

B

Q1 = 319 Q3 = 62

Q2 = 134

Q4 = 19 Q6 = 185

Q8 = 72 Q5 = 43

150

Q7 = 35

150 (c)

Fig. E11.6


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Solution There are five junctions (J 5), eight pipes (P 8), and two fixed-grade nodes (F 2). Hence the number of closed loops is L 8 5 2 1 2, plus one pseudoloop. The three loops and assumed flow directions are shown in Fig. E11.6b. Equation 11.4.26 is applied to loop I: ( W4 W3 W2 W1) (zA zB) QI G4 G3 G2 G1 Apply Eq. 11.4.27 to loops II and III: ( W2 W8 W7 W6) QII G2 G8 G7 G6 ( W3 W5 W8) QIII G3 G5 G8 The problem is solved using an Excel spreadsheet, and the solution is shown in Table E11.6. The spreadsheet layout showing the equations used in the solution is provided in Appendix F, Fig. F.6. In the loop equations note that W and G have the correct sign attributed automatically using the relations W R 苶Q앚Q앚 and G 2R 苶앚Q앚. In addition, the initial values of each Q take on a positive or negative sign depending on the assumed flow direction relative to the positive clockwise direction for each loop. The initial flow estimates, shown under the heading “Iteration 1,” are chosen to satisfy continuity. Note that flow adjustments to each pipe element are made after all of the Q’s have been computed; for example, relative to loop I, Q2 (Q0)2 Q1 QII. After four iterations the absolute magnitude of the Q’s are all less than 0.001, and the left-hand side of Eq. 11.4.29 is 0.0051. The values of Qi after four iterations are shown in Fig. E11.6c, along with the final flow directions. Flow rates are in liters per second. The piezometric heads are evaluated by computing the energy drop along designated paths, beginning at a known fixed-grade node, in this case, node A:

HC HA R1Q 12 50 100(0.319)2 39.8 m HD HC R2Q 22 39.8 500(0.134)2 30.8 m HE HD R3Q 23 30.8 200(0.062)2 30.0 m HF HC R6Q 62 39.8 300(0.185)2 29.5 m HG HD R8Q 82 30.8 300(0.072)2 29.2 m

Note that there is negligible loss in element four. (continued)


200 0.060 400 0.040 300 0.070

318.000

98.000

0.110

⌬Q 1.12E-03

24.000 0.064 32.000 0.041 42.000 0.074

0.720 0.640 1.470

⌬Q 4.87E-03

1.550

320.779

102.589

25.440 0.062 32.898 0.043 44.251 0.071

⌬Q 1.43E-03

0.146

0.809 0.676 1.632

226.081

3.899 24.817 133.466 63.899

2RⱍQⱍ

316.759

101.710

24.817 0.063 34.039 0.043 42.853 0.073

⌬Q 3.57E-04

0.036

0.770 0.724 1.530

20.000

RQⱍQⱍ

228.309

3.968 25.098 135.276 63.968

2RⱍQⱍ

317.183

102.941

25.098 0.062 34.325 0.043 43.519 0.072

⌬Q 52.29E-04

0.054

0.787 0.736 1.578

⌬Q 9.83E-05

0.031

0.019 0.062 0.134 0.319

Q

135.276 0.134 43.519 0.072 27.650 0.035 110.738 0.185

⌬Q 9.03E-04

0.206

133.466 0.135 9.150 42.853 0.073 1.578 28.823 0.035 0.478 111.617 0.185 10.219

⌬Q 1.47E-03

0.464

Q

0.020 0.039 0.063 0.787 0.135 9.150 0.320 10.230

⌬Q 3.44E-04

0.078

137.352 0.113 8.907 44.251 0.071 1.530 28.101 0.036 0.519 111.075 0.186 10.382

⌬Q 9.03E-04

0.290

130.000 0.137 9.433 42.000 0.074 1.632 32.000 0.035 0.494 114.000 0.185 10.281

A Mathcad solution for Example 11.6 is given in Appendix F, Fig. F.7.

Pipe3 Loop 3 Pipe5 Pipe8

500 0.130 8.450 300 0.070 1.470 400 0.040 0.640 300 0.190 10.830

⌬Q 2.48E-03 Pipe2 Loop 2 Pipe8 Pipe7 Pipe6

231.783

0.691

222.000

0.550

20.000

RQⱍQⱍ

0.019 0.038 0.062 0.770 0.133 8.907 0.319 10.208

Q

Iteration 4

11:26 AM

⌬Q 2.98E-03

4.495 25.440 137.352 64.495

20.000 0.022 0.051 0.064 0.809 0.137 9.433 0.322 10.399

4.000 24.000 130.000 64.000

20.000 0.020 0.040 0.060 0.720 0.130 8.450 0.320 10.240

100 200 500 100

⌬H

2RⱍQⱍ

RQⱍQⱍ

Q

2RⱍQⱍ

RQⱍQⱍ

Q

Iteration 3

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Pipe4 Loop 1 Pipe3 Pipe2 Pipe1

R

Iteration 2

570

Iteration 1

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11.4.4 Network Analysis Using Generalized Computer Software Spreadsheets and computational software such as Mathcad or MATLAB can be used to solve for the flow distribution in small networks, as illustrated in Example 11.6. However, for larger networks that contain more than several loops, the task of programming becomes cumbersome, especially since a unique description is required for each system accompanied by a large number of equations to solve. An advantage of using these methods of solution is that the user is applying the fundamental relations directly and can easily discern convergence of the solution. On the other hand, generalized computer codes have robust solution algorithms that allow a wide variety of network systems to be analyzed, and provide userfriendly input/output schemes. They have become an indispensable tool for many engineers and practitioners in the water-supply industry as well as finding use in other industrial applications. However, the user must become familiar with the code so that the correctness of the solution is assured. With the advancement of software development and the increase of personal computer memory and computational speed, it is now possible to analyze large, highly complex networks with relative ease. In addition to the Hardy Cross method outlined in Section 11.4.3, other linearized methods have been used in computer codes. Herein is described briefly the use of the program EPANET developed by the United States Environmental Protection Agency. EPANET is a comprehensive program that simulates both hydraulic flow and water quality in pressurized piping networks. For the hydraulic analysis, it uses a hybrid node-loop algorithm termed the gradient method to solve for the unknown piezometric heads and discharges. The network can contain pipes, pipe junctions, pumps, and various types of valves, reservoirs, and storage water tanks. Friction head loss is computed using the Darcy–Weisbach, Hazen–Williams, or Chezy–Manning formulas. Either pump curve data or useful power can be accommodated. Further capabilities of the program, a comprehensive description, and details of running the code are provided in the EPANET Users Manual (Rossman, 2000). The source code and users manual is available from the EPA Web site www.epa.gov.

Example 11.7 Rework Example 11.6 using EPANET. Solution The user-defined network map is shown in Fig. E11.7, and a partial listing of the computed results are provided in Table E11.7. For coding details refer to the EPANET Users Manual.

(continued)


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1

7

el = 50 m

4 2

1

2

7

3

6

8

6 Qe = 150 L/s

3

el = 30 m

4

5

5 Qe = 150 L/s

Fig. E11.7

Table E11.7 Link – Node Table: Link ID

1 2 3 4 5 6 7 8

Start Node

End Node

Length m

Diameter mm

1 2 3 4 4 3 2 6

2 3 4 7 5 5 6 5

66 330 130 66 260 200 200 260

250 250 250 250 250 250 250 250

Demand LPS

Head m

Pressure m

Quality

40.45 30.96 30.05 29.16 29.82 50.00 30.00

40.45 30.96 30.05 29.16 29.82 0.00 0.00

Node Results: Node ID

2 3 4 5 6 1 7

0.00 0.00 0.00 150.00 150.00 319.72 19.72

0.00 0.00 0.00 0.00 0.00 0.00 Reservoir 0.00 Reservoir

Link Results: Link ID

Flow LPS

Velocity m/s

Headloss m/km

Status

1 2 3 4 5 6 7 8

319.72 133.59 62.19 19.72 42.47 71.40 186.12 36.12

6.52 2.72 1.27 0.40 0.87 1.46 3.79 0.74

144.71 28.75 6.98 0.83 3.44 9.01 53.13 2.55

Open Open Open Open Open Open Open Open


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In addition to Example 11.7, solutions for the two networks shown in Fig. 11.7 are provided in Appendix F, Figs. F.8 and F.9. For the branching or treelike system shown in Fig. 11.7a, the Hazen–Williams friction formula is used with 40 –– K = 23

114; 50

;

P = number of pipe elements L = number of interior loops J = number of junctions F = number of fixed grade nodes Elevations of junctions are underlined

59

24 ––

82;

50

2;

35 ––

K=0

117; 100

27 ––

16

27 ––

5;

K = 23

10

0

K=2

50 26 ––

K = 23

33 ––

10

K = 5 (Loss coefficient) 188; 100 30 –– 130; 32 100 –– (Elev. in m) K = 10

195; 100 P 40.4 kW (Useful power)

K=0

10

30 ––

33 ––

64;

50

26 ––

26 –– 127; 50

(Length in m; diameter in mm)

155; 50

78; 100

35 ––

40 60; 50 ––

K = 23

158; 50

37 ––

K = 23

0

0

115; 50

38 ––

55; 5

K = 23

K = 23

K = 23

24 ––

K = 23

(a) 61 ––

305

0

0;

0; 4 152 55 L/s 55 L/s

100 L/s

8 16

44 –– 670; 380

43 ––

41 –– 760; 1

1380; 300

40 ––

35

100 L/s

1070; 300

46 –– 00

00

;4 1200

49 ––

; 300

0; 3

m)

50

168

. in

ev (El

K = 10

1100; 300

(Pump performance curve)

0; 3

00 46 12 350 –– ; 0 0 9

50

Q 0 L/s 850 1700

168

46 –– ;4 00 20

Hp 180 m 177 171

50

P

;3

50

20

15

50 –– 85 L/s 44 ––

140 L/s

3000; 600

15 ––

K=5

1520; 400

49 –– (Length in m; diameter in mm)

0

; 15

140 L/s

50

40 –– 55 L/s

(b)

Fig. 11.7 Two piping networks: (a) tree watering system: P 17, L 0, J 8, F 10; (b) water distribution system: P 17, L 4, J 12, F 2. (Wood, D. J. Algorithms for Pipe Network Analysis and Their Reliability, Research Report 127, Water Resources Research Institute, University of Kentucky, Lexington, KY., 1981)


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the coefficient C 130. The useful input power of the pump is 40.4 kW. For the looping network shown in Fig. 11.7b, friction losses are based on the Darcy–Weisbach relation with an absolute roughness of 0.15 mm for all pipes and a kinematic viscosity of 106 m2앛s. The pump performance curve is defined by three settings of pump head and discharge.

11.5 UNSTEADY FLOW IN PIPELINES KEY CONCEPT For unsteady flow to occur in piping, an excitation to the system is required (e.g., sudden closure of a valve).

Unsteady, or transient flows in pipelines traditionally have been associated with hydropower piping and with long water and oil pipeline delivery systems. However, the application has broadened in recent years to include hydraulic control system operation, events that take place in power plant piping networks, fluid–structure interaction in liquid-filled piping, and pulsatile blood flow. For the unsteadiness to occur, some type of excitation to the system is necessary. Representative excitations are valve opening or closure, pump or turbine operations, pipe rupture or break, and cavitation events. In this section we examine only the most fundamental aspects of unsteady flow in piping. First we study unsteady flow in a single constant-diameter pipe assuming inelastic and incompressible conditions; that is followed by analysis of a system in which elasticity and compressibility play a significant role in the response of the pressure and velocity to an excitation. The first assumption results in the phenomenon called surging, while the second one leads to the phenomenon called water hammer.

11.5.1 Incompressible Flow in an Inelastic Pipe Consider a single horizontal pipe of length L and diameter D (Fig. 11.8). The upstream end of the pipe is connected to a reservoir and a valve is situated at the downstream end.The upstream piezometric head is H1, and downstream of the valve the piezometric head is H3. Note that both H1 and H3 are constant time-independent reservoir elevations. The friction factor f is assumed constant, and the loss coefficient for the valve is K. There are a number of possible excitations that could be considered, but we will only look at the situation where initially there is a steady velocity V0, the valve is then instantaneously opened to a new position, and

H1 H3 V 1

4

2

3

L

Fig. 11.8

Horizontal pipe with a valve at the downstream end.


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Sec. 11.5 / Unsteady Flow in Pipelines 575

subsequently the flow accelerates, increasing to a new steady-state velocity Vss. For situations in which the valve is closed, either partially or completely, one should consider the possibility of water hammer taking place; this is treated in the next section. Here, we assume that the velocity does not vary with position, only with time. In Fig. 11.8 define a control volume for the liquid in the pipe between locations 1 and 2, whose mass is rAL. Note that location 2 is upstream of the valve. At the two ends of the control volume the pressures are p1 and p2, respectively, and on the surface the wall shear stress is t0. The conservation of momentum for that liquid volume is given by dV A(p1 p2) t0pDL rAL dt

KEY CONCEPT The velocity does not vary with position, only with time.

(11.5.1)

Without serious consequences we can assume steady-flow conditions across the valve from location 2 to 3, and utilize the energy equation to provide rV 2 p2 p3 K 2

(11.5.2)

It is reasonable to assume that the Darcy–Weisbach friction factor based on steady-state flow can be employed without undue error. Then, from Section 7.3.3, Eq. 7.3.19, we have the relation rf V 2 t0 8

(11.5.3)

Substituting Eqs. 11.5.2 and 11.5.3 into Eq. 11.5.1, dividing by the mass of the liquid column, and recognizing that p1 p3 rg(H1 H3), since the velocity heads are assumed to be negligible, there results H1 H3 f dV K V2 g 0 L dt D L 2

冢

冣

(11.5.4)

Equation 11.5.4 is the relation that represents incompressible unsteady flow in the pipe. The initial condition at t 0 is a given velocity V V0. When the final steady-state condition is attained, dV/dt 0, and that steady-state velocity, designated as Vss, can be obtained by setting the derivative to zero in Eq. 11.5.4:

Vss

莦冪莦 fL앛D莦 K 2g(H1 H3)

(11.5.5)

Substituting Eq. 11.5.5 into Eq. 11.5.4, separating variables, and expressing the result in integral form, one has

冕

t

V 2ssL dt g(H1 H3) 0

冕

V

dV 2 V2 V ss 0

(11.5.6)

V


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Upon integration, the resulting relation defines the velocity V relative to time t after the valve excitation: (Vss V)(Vss V0) V L t ss ln (Vss V)(Vss V0) 2g(H1 H3)

(11.5.7)

There are some significant features to the solution. First, by studying Eq. 11.5.7 one concludes that infinite time is required to reach the steady-state velocity Vss. In reality, Vss will be reached some time sooner, because the losses have not been completely accounted for. However, it is possible to determine the time when a percentage, for example 99%, of Vss has been reached that would be adequate for engineering purposes. In addition, it is possible for the fluid to be initially at rest, that is, V0 0. Finally, remember that the solution is based on the assumptions that both liquid compressibility and pipe elasticity are ignored; the next section addresses the situation when those assumptions are invalid.

Example 11.10 A horizontal pipe 1000 m in length, with a diameter of 500 mm, and a steady velocity of 0.5 m/s, is suddenly subjected to a new piezometric head differential of 20 m when the downstream valve suddenly opens and its coefficient changes to K 0.2. Assuming a friction factor of f 0.02, determine the final steady-state velocity, and the time when the actual velocity is 75% of the final value. Solution Given are L 1000 m, D 0.5 m, V0 0.5 m/s, f 0.02, and K 0.2. Setting H1 H3 20 m, the final steady-state velocity Vss is found by direct substitution into Eq. 11.5.5: Vss

2 9.81 20 3.12 m앛s 冪莦莦 0.02 1000앛0.5 0.2

The velocity that is 75% of Vss is V 0.75 3.12 2.34 m/s. Then, using Eq. 11.5.7, the time corresponding to that velocity is (3.12 2.34) (3.12 0.5) 3.12 1000 t ln 12.9 s (3.12 2.34) (3.12 0.5) 2 9.81 20 Hence the final steady-state velocity is 3.12 m/s, and the time when 75% of that velocity is attained is approximately 13 s.

11.5.2 Compressible Flow in an Elastic Pipe In contrast to the developments of Section 11.5.1, there are situations where the liquid is not incompressible and the piping is not rigid. Rather, the interaction between changes in momentum and applied forces causes the liquid to slightly compress and the pipe material to experience very small deformations. When this


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occurs, significant pressure changes can take place, and the phenomenon is termed water hammer. Water hammer is accompanied by pressure and velocity perturbations traveling at very high velocities, close to the speed of sound in the liquid. The resulting wave action occurs at relatively high frequencies. We will consider the situation in which a valve at the downstream end of a pipe closes or opens suddenly, either partially or completely, to initiate a water hammer response. First let us develop the fundamental equations. Consider again the horizontal pipe shown in Fig. 11.8, where now the valve will close so rapidly that elastic effects cause water hammer to occur. The movement of the valve will cause an acoustic, or pressure, wave with speed a to propagate upstream. A control volume of an incremental section of liquid contained in the pipe is shown in Fig. 11.9a, where the pressure wave at a given instant is located. The presence of the wave implies that an unsteady flow is taking place within the control volume; at the entrance the velocity is V, and at the exit it is V V. Steady-state conservation laws can be applied if the wavefront is made to appear stationary by an observer moving with the wavespeed (Fig. 11.9b). (Refer to Section 4.5.3 for the development of inertial reference frames that move with a constant velocity.) In contrast to Fig. 11.9a, the entrance velocity is now V a, and at the exit it is V V a. The pressure, pipe area, and density at the entrance are p, A, and r, respectively. Due to the passage of the pressure wave, at the exit the pressure, pipe area, and

KEY CONCEPT Water hammer is accompanied by pressure and velocity perturbations that travel at high velocities.

Instantaneous position of wave

V

V + ΔV

a

(a)

V + ΔV + a

V+a

(b)

pA

(p + Δ p)(A + Δ A)

(p + Δ p) Δ A

(c)

Fig. 11.9 Pressure wave moving through a segment of horizontal pipe: (a) pressure wave moving to the left at speed a; (b) pressure wave appears stationary using the principle of superposition; (c) pressure forces acting on the control volume.


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liquid density are altered to p p, A A, and r r, where p, A, and r are the respective changes in pressure, area, and density. Applying the conservation of mass across the control volume in Fig. 11.9b, we find that 0 (r r)(V V a)(A A) r(V a)A

(11.5.8)

Neglecting frictional and gravitational forces, only pressure forces act on the control volume in the direction of flow, as shown in Fig. 11.9c. The conservation of momentum across the control volume is pA ( p p) A ( p p)(A A) rA(V a)[V V a (V a)]

(11.5.9)

Equations 11.5.8 and 11.5.9 are expanded, and the terms containing factors of 2 and 3 are dropped, since they are much smaller in magnitude than the remaining ones. Then Eqs. 11.5.8 and 11.5.9 become rA V (V a)(A r r A) 0

(11.5.10)

A p rA(V a) V

(11.5.11)

and

In nearly all situations V a, so that Eq. 11.5.11 becomes p ra V Joukowsky equation: Relates the pressure change to the pressure wave speed and the change in velocity.

(11.5.12)

Equation 11.5.12, called the Joukowsky equation, relates the pressure change to the pressure wave speed and the change in velocity. Note that a velocity reduction (a negative V) produces a pressure rise (a positive p), and a positive V yields a negative p. Once the wave has passed through the control volume, the altered conditions p p, V V, A A, and r r will persist until the wave reflects from the upstream boundary; this wave action will be discussed later. First it is useful to examine the nature of the pressure pulse wave speed a. Equations 11.5.10 and 11.5.11 are combined, again recognizing that V a, to eliminate V, with the result that r A p 2 r A ra

(11.5.13)


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From the definition of the bulk modulus of elasticity B for a fluid, Eq. 1.5.11, we can relate the change in density to the change in pressure as r앛r p앛B. The change in pipe area can be related to the change in pressure by considering an instantaneous, elastic response of the pipe wall to pressure changes. Assuming a circular pipe cross section of radius r, we have A/A 2 r/r, and the change in circumferential strain in the pipe wall is r/r. For a thin-walled pipe whose thickness e is much smaller than the radius, that is, e r, the circumferential stress is given by s pr앛e. For small changes in r and e, s ⬇ (r앛e) p. The elastic modulus for the pipe wall material is the change in stress divided by the change in strain, or s (r앛e) p (2r 앛e) p E ⬇ r앛r A앛A

(11.5.14)

Solving for A/A, and substituting the result into Eq. 11.5.13, along with the relative change in density related to the change in pressure and bulk modulus, there results p 2r p r 2 B E ra

(11.5.15)

The parameter p can be eliminated in Eq. 11.5.15, the diameter substituted for the radius, and the relation solved for a:

a

冪莦 앛e)(B莦 1 (D莦 앛E) B앛r

(11.5.16)

Hence the pressure pulse wave speed is shown to be related to the properties of the liquid (r and B) and those of the pipe wall (D, e, and E). If the pipe is very rigid, or stiff, then the term DB/eE 1, and Eq. 11.5.16 becomes a 兹B앛r 苶, which is the speed of sound in an unbounded liquid (see Eq. 1.5.12). Note that the effect of pipe elasticity is to reduce the speed of the pressure wave. The use of Eqs. 11.5.12 and 11.5.16 will provide only the magnitude of the water hammer excitation. In addition, it is necessary to understand the nature of the pressure pulse wave as it travels throughout the pipe and is reflected from the boundaries. Consistent with the development of Eqs. 11.5.12 and 11.5.16, we neglect friction to simplify the analysis. Consider the situation in which the valve is suddenly closed and the pipe is rigid. Let us study in detail the sequence of events throughout one cycle of motion, as illustrated in Fig. 11.10. In Fig. 11.10a an initial steady condition exists, the velocity is V0, and the valve is suddenly closed at t 0. Note that when friction is neglected, the hydraulic grade line appears horizontal. Subsequent to closure of the valve, the wave travels upstream (Fig. 11.10b). Behind the wave, the velocity is reduced to zero, the pressure rises by the amount p, the liquid has been compressed, and the pipe slightly expanded. At time L/a the wave reaches the reservoir, and an


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a

HGL

Δp

a

H1 V0

–V 0 (d) t = 3L/2a

(a) t = 0 Location of wave

V=0 (g) t = 3L/a

HGL Δp

a

Δp

V0

–V 0

V=0

(b) t = L/2a

a

V0 (h) t = 7L/2a

(e) t = 2L/a

HGL Δp

V=0 (c) t = L/a

a

–V 0

V=0

(f) t = 5L/2a

V0 1

4

2

(i) t = 4L/a

Fig. 11.10 One cycle of wave motion in a pipe due to a sudden valve closure.

unbalanced force occurs at the pipe inlet (Fig. 11.10c). In the pipe the pressure reduces to the reservoir pressure, and the velocity suddenly reverses direction; this process begins at the upstream end and propagates downstream at the wave speed a (Fig. 11.10d). When the wave reaches the valve at time 2L/a, the velocity has magnitude V0 throughout the pipe (Fig. 11.10e). Adjacent to the valve, which is now closed, the velocity reduces to zero, and the pressure reduces by the amount p (Fig. 11.10f). The low-pressure wave travels upstream at speed a, and behind the wave the liquid is expanded and the pipe wall is contracted. Note that if the pressure behind the wave reduces to vapor pressure, cavitation will occur, and some of the liquid will vaporize. When the pressure wave reaches the reservoir at time 3L/a, an unbalanced condition occurs again, opposite in magnitude to that at time L/a (Fig. 11.10g). Equilibrium of forces will now cause a wave to travel downstream, with the pressure increased by the amount p and the velocity equal to V0 behind the front (Fig. 11.10h). When the wave reaches the valve at time 4L/a, initial steady-state conditions prevail once again throughout the pipe (Fig. 11.10i). The process repeats itself every 4L/a seconds. For the ideal, frictionless situation shown here, the motion will perpetuate. The pressure waveform at the valve and at the midpoint of the pipe and the velocity at the pipe entrance are shown in Fig. 11.11. For a real pipe system, friction, pipe motion, and inelastic behavior of the pipe material will eventually cause the oscillation to die down and cease (see Fig. 11.12).


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Sec. 11.5 / Unsteady Flow in Pipelines 581 p2 Δ p = ρ aV0

p1 = γ H1 0

0

2

4

6

8

10

12

14

16

18

20

2

4

6

8

10

12

14

16

18

20

2

4

6

8

10

12

14

16

18

20

ta/L

p4 Δ p = ρ aV0

p1 = γ H1 0

0

ta/L

V1

V0 0

0

ta/L

Fig. 11.11 Pressure waveforms at the valve (p2), pipe midpoint (p4), and velocity waveform at the entrance to the pipe (V1).

The pressure rise p predicted by Eq. 11.5.13 is based on the assumption that the valve closes instantaneously. Actually, it can be used to predict the maximum pressure rise when the valve closes with any time less than 2L/a, the time it takes for the pressure wave to travel from the valve to the reservoir and back again. For valve closure times greater than 2L/a, a more comprehensive analysis is required (Wylie and Streeter, 1993). Keep in mind that the previous discussion

Absolute pressure (meters of water)

200

100

0

0

200

400

600

800

Time (ms)

Fig. 11.12 Pressure waveform at the valve for an actual pipe system following rapid valve closure. (After Martin, 1983.)(Martin, C. D., Experimental Investigation of Column Separation with Rapid Valve Closure, Proceedings, 4th International Conference on Pressure Surges, BHRA Fluid Engineering, Cranfield, England, 1983, pp. 77–88. Reproduced with permission.)


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pertains only to a single horizontal pipe with a reservoir at the upstream end and a valve closing instantaneously at the downstream end and one that contains a frictionless liquid. Most of the water hammer analysis performed by engineers today makes use of computer-based numerical methods for complex piping systems, incorporating a variety of excitation mechanisms such as pumps, surge suppressers, and various types of valves (Wylie and Streeter, 1993).

Example 11.11 A steel pipe (E 207 MPa, L 1500 m, D 300 mm, e 10 mm) conveys water at 20°C. The initial velocity is V0 1 m/s. A valve at the downstream end is closed so rapidly that the motion is considered to be instantaneous, reducing the velocity to zero. Determine the pressure pulse wave speed in the pipe, the speed of sound in an unbounded water medium, the pressure rise at the valve, the time it takes for the wave to travel from the valve to the reservoir at the upstream end, and the period of oscillation. Solution The density and bulk modulus of water at 20°C are found in Table B.1: r 998 kg/m3 and B 220 107 Pa. The pressure wave speed, a, is given by Eq. 11.15.16: a

220 10 앛998 1290 m앛s 莦莦 220 10 冪 1 0.3 7

7

0.01

207 109

The speed of sound in an unbounded water medium is found by use of Eq. 1.5.12 to be a

220 10 1485 m앛s 冪莦 998莦 7

Note that the sound speed is about 15% larger than the pressure wave speed. To compute the pressure rise at the valve, we recognize that the reduction in velocity is V0 1 m/s. Using Eq. 11.15.12, the increase in pressure upstream of the valve is p 998 1290 ( 1) 1.29 106 Pa or 1290 kPa The wave travel time from the valve to the reservoir is L/a 1500/1290 1.16 s, and the period of oscillation is 4L/a 4 1.16 4.65 s.

11.6 SUMMARY A methodology for dealing with losses in piping has been provided in Section 11.2 using the resistance coefficient as a generalized term that can include minor losses along with a chosen empirical friction loss. For accurate representation of losses, the Darcy–Weisbach friction loss is recommended, with losses proportional to the square of the velocity (or discharge). Equation 11.2.3 is useful for estimating the Darcy–Weisbach friction factor in an iterative, or trial, solution. In


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Problems 583

Section 11.3 simple pipe systems have been defined as those that contain one to several pipes, arranged singly, in series, parallel, or branched. Typically, a problem consists of finding the flow distribution in the piping; along with discharges the piezometric head often is an unknown. Calculator-based solutions or computational software are useful tools to analyze these systems. For more complex piping networks, a systematic solution approach has been presented in Section 11.4. First the network equations were linearized, and then the Hardy Cross method was applied to solve for the network flows and piezometric heads. A spreadsheet solution was applied to a network using the Hardy Cross solution; it allows one to correctly apply the energy and continuity equations as well as track the convergence of the solution. Note that even the simple pipe systems presented in Section 11.3 can be solved using the Hardy Cross method. The computer program EPANET was also introduced in Section 11.4. This software is useful for large systems in which spreadsheet programming becomes cumbersome. Finally, analysis of unsteady flows in pipes has been introduced in Section 11.5. We have focused on flow that behaves either in an incompressible or compressible manner; the latter behavior is commonly called “water hammer.” Water hammer is a result of acoustic waves propagating in a pipe. The waves travel at the speed of sound in the contained liquid and produce a periodic response of both pressures and velocities throughout the pipe. The Joukowsky relation, Eq. 11.5.12, gives the magnitude of the pressure wave.

PROBLEMS Steady Flows 11.1

11.2

Using the results from Example 11.2, tabulate and plot to correct vertical scale the hydraulic and energy gradelines. Assume a 10-m length between the two elbows. A pump is situated between two sections in a horizontal pipeline. The diameter D1 and pressure p1 are given at the upstream section, and D2 and p2 are given at the downstream section. Determine the required fluid power of the pump for the following conditions: (a) D1 50 mm, p1 350 kPa, D2 80 mm, p2 760 kPa, Q 95 L/min, hL 6.6 m, water is flowing at 20°C. (b) D1 2 in, p1 50 lb/in2, D2 3 in, p2 110 lb/in2, Q 25 gal/min, hL 20 ft, water is flowing at 70°F.

11.3

An oil (S 0.82) is pumped between two storage tanks in a pipe with the following characteristics: L 2440 m, D 200 mm, f 0.02, K 12.5. The upper tank is 32 m higher than the lower one. Using the pump data provided, determine: (a) The oil discharge in the pipe. (b) The power requirement for the pump.

Q (L앛s)

0

15

30

45

60

75

100

HP (m)

55

54

53

52

49

44

35

h

0

0.4

0.6

0.7

0.75

0.7

0.5


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11.4

Consider the simple pumping system shown in Fig. P11.4. Assume the following parameters to be known:

(b) What is the maximum elevation possible at location C without causing vapor pressure conditions to exist? (c) Sketch the hydraulic grade line.

Reservoir elevations, z1, z2 Pipe length, L Pipe roughness, e Sum of minor loss coefficients, K Kinematic viscosity, n Discharge, Q

C e = 1 mm, D = 500 mm

Develop a numerical solution to find the required diameter. Use an equivalent length to represent the minor losses. Determine the diameter for the following data: (a) z2 z1 120 ft, L 1500 ft, e 0.003 ft, K 2.5, water at 50°F, Q 15,000 gal/min, pump characteristic curve: HP (ft) Q (gal앛min)

490

486

473

453

423

406

0

4000

8000

12,000

16,000

18,000

160

158

154

148

138

Q (L앛s)

0

400

800

1200

1600

2

1 km

elB 220 m B

4 km A

P

Fig. P11.5 11.6

For the three pipes in series shown in Fig. P11.6, minor losses are proportional to the discharge squared, and the Hazen–Williams formula is used to account for friction losses. With the data given, use Newton’s method to determine the discharge. Note that minor losses can be neglected in order to initially estimate Q. (a) (p/g z)A 250 m and (p/g z)B 107 m.

Pipe

L (m)

D (mm)

K

C (Hazen–Williams)

1 2 3

200 150 300

200 250 300

2 3 0

100 120 90

(b) (p/g z)A 820 ft and (p/g z)B 351 ft.

1 P

Fig. P11.4 11.5

Kv = 2

elA 100 m

(b) z2 z1 40 m, L 500 m, e 1 mm, K 2.5, water at 20°C, Q 1.1 m3/s, pump characteristic curve: HP (m)

Kv = 2

Gasoline is being pumped at 400 L/s in a pipeline from location A to B, as shown in Fig. P11.5. The pipe follows the topography as shown, with the highest elevation shown at location C. The only contributions to minor losses are the two valves located at the ends of the pipe. If S 0.81, n 4.26 107 m2/s, pv 55.2 kPa absolute, and patm 100 kPa, (a) Determine the necessary power to be delivered to the system to meet the flow requirement.

Pipe

L (ft)

D (in.)

K

C (Hazen–Williams)

1 2 3

600 300 900

8 10 12

2 3 0

100 120 90

A [2] [1]

[3]

B

Fig. P11.6


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Problems 585 11.7

A long oil pipeline consists of the three segments shown in Fig. P11.7. Each segment has a booster pump that is used primarily to overcome pipe friction. The two reservoirs are at the same elevation. (a) Derive a single equation to determine the discharge in the system if the resistance factors R 苶i for each pipe and the useful power · Wfi for each pump are known. (b) Determine the discharge for the data given in the figure. The unit weight of oil is g 8830 N/m3. Q – – R3 = 2 × 105 s2/m5 R1 = 4 × 104 s2/m5 – R2 = 3 × 104 s2/m5 P1 P2 P3

A

Wf1 = 200 kW

Wf2 = 200 kW

B

Wf3 = 200 kW

Fig. P11.7 11.8

A liquid with a specific gravity of 0.68 is pumped from a storage tank to a free jet discharge through a pipe of length L and diameter D (Fig. P11.8). The pump provides a known amount of fluid · power Wf to the liquid. Assuming a constant friction factor of 0.015, determine the discharge for the following conditions: (a) z1 24 m, p1 110 kPa, z2 18 m, · L 450 m, D 300 mm, Wf 10 kW. 2 (b) z1 75 ft, p1 15 lb/in , z2 60 ft, · L 1500 ft, D 8 in, Wf 15 hp.

el 50 m

Fig. P11.9 11.10

Find the water flow distribution in the parallel system shown in Fig. P11.10, and the required pumping power if the discharge through the pump is Q1 3 m3/s. The pump efficiency is 0.75. Assume constant friction factors.

Pipe

L (m)

D (mm)

f

K

1 2 3 4

100 1000 1500 800

1200 1000 500 750

0.015 0.020 0.018 0.021

2 3 2 4

Kelbow = 0.26

Kv = 2

Fig. P11.8 11.9

Water at 20°C is being pumped through the three pipes in series as shown in Fig. P11.9. The power delivered to the pump is 1920 kW, and the pump efficiency is 0.82. Compute the discharge.

B A

[3]

P [1]

C

[4]

Fig. P11.10 11.11

For the system shown in Fig. P11.11, determine the water flow distribution and the piezometric head at the junction using an ad hoc approach. Assume constant friction factors. The pump characteristic curve is HP a bQ2. (a) a 20 m, b 30 s2/m5, z1 10 m, z2 20 m, z3 18 m.

z2

Ke = 0.5

el 20 m

[2] el 0 m

z1 Kv = 2

[1] P

p1

P

[3]

[2]

el 0 m

Pipe

L (m)

D (cm)

f

K

1 2 3

30 60 90

24 20 16

0.020 0.015 0.025

2 0 0

(b) a 55 ft, b 0.1 sec2/ft5, z1 20 ft, z2 50 ft, z3 45 ft.

Pipe

L (m)

D (mm)

e (mm)

K

Pipe

L (ft)

D (in.)

f

K

1 2 3

200 300 120

1500 1000 1200

1 1 1

2 0 10

1 2 3

100 200 300

10 8 6

0.020 0.015 0.025

2 0 0


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Chapter 11 / Flows in Piping Systems p0

2

3

[2] [3]

[2]

1

[3]

[1]

P

[1] [4]

Fig. P11.14

Fig. P11.11 11.12

Work Problem 11.11 using a numerical approach: (a) Newton’s method. (b) False position method. 11.13 Find the flow distribution in the parallel network shown in Fig. P11.13. Assume constant friction factors. The change in hydraulic grade line between A and B is (p/g z)A (p/g z)B 50 m. Pipe

L (m)

D (mm)

e (mm)

K

1 2 3 4

600 1000 550 800

1000 1200 850 1000

0.1 0.15 0.2 0.1

2 0 4 1

11.15

For the system shown in Fig. P11.15, a fluid flows from A to B. Determine in which pipe there is the greatest velocity. Pipe

L (m)

D (mm)

f

1 2 3

2000 650 1650

450 150 300

0.012 0.020 0.015

[1] [2] B

A [2]

Q

[3]

Fig. P11.15

[1] A

11.16

[3] [4]

B

Fig. P11.13 11.14

The water sprinkling system shown in Fig. P11.14 is applied from a large diameter pipe with constant internal pressure p0 300 kPa. The system is positioned in a horizontal plane. Determine the flow distribution Q1, Q2, Q3, Q4 for the given data. Valve losses are included in the 苶 R, values. (Hint: If you are clever, no trial-and-error solution is necessary!) Pipe

R (s2앛m5) 苶

1 2 3 4

1.6 104 5.3 105 1.0 106 1.8 106

A proposed water irrigation system consists of one main pipe with a pump and three pipe branches (Fig. P11.16). Each branch is terminated by an orifice, and each orifice has the same elevation. It is apparent that the flow distribution can be solved by treating the piping arrangement as a branched system. However, the piping can also be treated as a parallel system in order to determine the flows. (a) Identify the equations and unknowns to satisfy the solution for a parallel system. Why is it possible to treat the irrigation system as a parallel piping problem? (b) Why would the solution to a parallel system be preferred to that of a branching system? (c) Determine the flow distribution and sketch the hydraulic grade line. (d) What part of the piping would one change to approximately double the discharge, assuming that the individual lengths and pump curve could not be altered?


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Problems 587 – R2 = 82 500 – R3 = 127 900 – R4 = 115 500

– R1 = 34 650 el 0 m

P

11.19 el 0 m

Hp = 45 – 1 × 104Q2 – – (Hp in m; Q in m3/s; R in s2/m5, orifice loss included in R)

Water is being pumped in the piping system shown in Fig. P11.19. The pump curve is approximated by the relation HP 150 5Q 21, with HP in meters and Q1 in m3/s. The pump efficiency is h 0.75. Compute the flow distribution and find the required pump power.

Fig. P11.16 11.17 Determine the required head (HP) and discharge (Q1) of water to be handled by the pump for the system shown in Fig. P11.17. The discharge in pipe 2 is Q2 35 L/s in the direction shown.

Pipe

R (s2앛m5) 苶

1 2 3

400 1000 1500

el 15 m el 12 m

el 40 m

– R2 = 2000

[2]

P

el 10 m

– R3 = 1500

Q2

el 3 m

el 10 m

– R4 = 1000

– R1 = 1400 s2/m5

P

[3] [1]

Fig. P11.17 11.18

Fig. P11.19

Determine the flow distribution of water in the parallel piping system shown in Fig. P11.18. (a) Qin 600 L/min Pipe

L (m)

D (mm)

f

K

1 2 3

30 40 60

50 75 60

0.020 0.025 0.022

3 5 1

11.20 A pipeline consists of two pipe segments in series (Fig. P11.20). The specific gravity of the fluid is 0.81. If pump A has a constant power input of 1 MW, find the discharge, the pressure head in pumps A and B, and the required power for pump B. The minimum allowable pressure on the suction side of pump B is 150 kPa, and both pumps have an efficiency of 0.76.

(b) Qin 0.35 ft3/sec Pipe

L (ft)

D (in.)

f

K

Pipe

L (m)

D (mm)

K

f

1 2 3

90 120 180

2 3 2.5

0.020 0.025 0.022

3 5 1

1 2

5000 7500

750 750

2 10

0.023 0.023

el 50 m

[1] el 27 m [2]

Qin

el 0 m [1] P

[3]

[2]

P B

A

Fig. P11.18

Fig. P11.20


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Chapter 11 / Flows in Piping Systems Determine the flow distribution of water in the system shown in Fig. P11.21. The equivalent roughness for all elements is 0.1 mm. Pipe

L (m)

D (mm)

K

1 2 3 4 5 6

1000 200 250 340 420 500

200 25 25 30 40 175

3 0 0 2 0 5

11.23

Determine the flow distribution of water in the system shown in Fig. P11.23. Assume constant friction factors, with f 0.02. The head-discharge relation for the pump is HP 60 10Q2, where HP is in meters and the discharge is in cubic meters per second. Pipe

L (m)

D (mm)

K

1 2 3 4 5

100 750 850 500 350

350 200 200 200 250

2 0 0 2 2

el 70 m

el 50 m [1]

[2]

el 48 m [4]

[2] [3] el 10 m

[4] [6]

[5]

[5]

el 0 m P [1]

Fig. P11.21 11.22

[3]

Fig. P11.23

Determine the flow distribution of water in the branching pipe system shown in Fig. P11.22: (a) Without a pump in line 1. (b) Include a pump located in line 1 adjacent to the lower reservoir. The pump characteristic curve is given by HP 250 0.4Q 0.1Q2, with HP in meters and Q in m3/s. The Hazen–Williams coefficient C 130 for all pipes. Pipe

L (m)

D (mm)

1 2 3 4

200 600 1500 1500

500 300 300 400

el 300 m el 250 m

11.24

A pump, whose performance and efficiency curves are given in Fig. P11.24a has been selected to deliver water in a piping system. The piping consists of four pipes arranged as shown in Fig. P11.24b. Water at 60°F is being pumped from reservoir A and exits at either reservoir B or at location D, depending on whether the valves at those locations are open or closed. The pipe characteristics are shown in the accompanying table. All of the pipe diameters are 4 in., and the friction factor in each pipe is assumed to be f 0.04. (a) If the discharge through the pump is 5,000 gal/hr, what is the head loss across pipe 2? (b) Compute the discharge in the system, assuming that the valve at location D is closed. (c) If the valve at location D is open and the discharge through the pump is 11,000 gal/hr, determine the discharge in pipe 4.

[3] [2]

el 30 m

[1]

el 200 m

[4]

Fig. P11.22

Pipe

L (ft)

K

1 2 3 4

10 500 2000 750

1 2 2 4


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Problems 589

HP (ft)

500 480

0.9

460

0.8

440

0.7

420

0.6

η

400 0

5000

10,000 Q (gal/hr)

15,000

20,000

(a) el 430 ft el 300 ft

B

[3]

[2] [1] P

C [4]

el 0 ft A

D

el 445 ft

(b)

Fig. P11.24

11.25

Refer to the system and pump curves associated with Problem 11.24. Assume that the discharge through the pump is 12,000 gal/hr. (a) What is the required power for the pump? (b) Determine the pressure at location C. (c) If the valve at location D is closed, what is the gage pressure at that location?

11.26

Refer to the system and pump curves associated with Problem 11.24. (a) Assuming that the valve at location D is open and the head rise across the pump is 460 ft, determine the discharge in pipe 3. (b) If the valve at location B was closed and the valve at location D was open, what is the head rise across the pump?


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11.27

For the network shown in Fig. P11.27, perform the following: (a) Identify all of the unknown parameters and write the required system equations. Subsequently, reduce them to one equation in one unknown. Assume b 2. (b) Prepare the network for a Hardy Cross analysis by making a sketch, identifying necessary loops and paths, and writing the appropriate equations.

11.28

The branching system shown in Fig. P11.28a is to be solved by using the Hardy Cross method. (a) Write the equations for QI, QII, W, and G. (b) Perform two iterations by filling in the blanks in the table shown in Fig. P11.28b. After two iterations, sketch the hydraulic grade line. el 50 el 40

I [1]

A

II

[2]

el 25

[2] [1] [3]

[3]

B (a)

J [4]

Loop

Pipe

– R

– 2R

Q

1

2

4

–3.5

2

2

4

0

W

G

Q

W

G

Q

[5] I C

Fig. P11.27 Δ QI =

2

2

4

0

3

2

4

–3.5

ΔQI =

II

ΔQII =

ΔQII =

(b)

Fig. P11.28

11.29

Determine the flow and the hydraulic grade line in the system shown in Fig. P11.29 using the Hardy Cross analysis method. Use g 9800 N/m3.

el 25 m

el 0 m R2 = 20 s2/m5

el 5 m

Q1

J

P R1 = 30 s2/m5

•

Wf = 200 kW

Q0 = 0.5 m3/s

Fig. P11.29


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Problems 591 11.30

The solution of flow of water in a pipe network is shown in Fig. P11.30. Compute: (a) The hydraulic grade line throughout the system. (b) The pressure at each node.

11.33

Determine the water flow distribution in the piping system shown in Fig. P11.33 and the pressures at the internal nodes employing the Hardy Cross method of solution. Pipe

L (m)

D (mm)

e (mm)

K

1 2 3 4 5

500 600 50 200 200

300 250 150 250 300

0.15 0.15 0.15 0.15 0.15

0 0 10 2 2

50 L /s – R5 = 310

el 0 m

200 L /s

0

Q4

=5

L/s

B

Q3 = 170 L/s

el 5 m C

– R3 = 280 s2/m5

200 L /s

– R1 = 20

51

[6]

el 100 m

el 20 m

␤= 2

– 2= R

D

A el 10 m

el 15 m [1]

Fig. P11.30

el 4 m [4]

A

el 0 m

11.31

Elevation Reservoir (ft) A B C

11.32

650 575 180

75 L/s

[3]

For the configuration shown in Problem 11.27, determine the water flow distribution and the hydraulic grade line at J. The system data are tabulated below.

el 2 m

[5]

[2] el 1 m

B

Fig. P11.33 Line

L (ft)

D (in.)

f

K

1 2 3 4 5

800 600 650 425 1000

8 3 3 3 4

0.015 0.020 0.020 0.025 0.015

0 2 2 3 4

Water is flowing in the piping system shown in Fig. P11.32. Determine the flow distribution using the Hardy Cross method:

11.34

Solve Problem 11.9 by employing the Hardy Cross method. Use Eq. 11.2.6 to compute the friction factors. 11.35 Determine the discharge in the piping system shown in Fig. P11.35: (a) With an exact solution. (b) Using the Hardy Cross method. Water is the flowing liquid and the losses are proportional to the velocity squared (b 2). The pump characteristic curve is HP 100 826Q2 (HP in meters and Q in m3/s). el 35 m – R1

– R3 = 2 A Qe = 35

␤= 2

– R1 = 3

=5

000

s 2/m

5

el 10 m – R2 = 300 s2/m5

– R2 = 5 s2/m5

P

B

Fig. P11.35 Qe = 15 L/s

Fig. P11.32


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Line

L (m)

D (mm)

e (mm)

K

1 2 3 4 5

200 150 500 35 120

100 50 100 50 100

0.1 0.1 0.1 0.1 0.1

2 30 0 35 2

el 125 m

Derive an equivalent single-pipe resistance coefficient R2 for the condenser tubes. (b) Write the energy equation for the entire system, using the required given variables and expressing flow resistance in terms of the resistance coefficients for pipes 1 and 3, and the equivalent resistance coefficient for pipe 2. (c) Set up a Hardy Cross algorithm to determine the discharge through the system. (d) If zA 2 m, zB 0 m, L1 100 m, D1 2 m, L2 15 m, D2 0.025 m, N 1000, L3 200 m, D3 2 m, and f 0.02, compute the discharge and hydraulic grade line. The coefficients for the pump curve are a0 30.4, a1 31.8, a2 18.6, and a3 4.0, where Q is in cubic meters per second. (e) If the top of the condenser is situated at an elevation of 6 m, what are the pressures at the top of the upstream and downstream headers?

el 118 m [2]

[1] A

C B D

[3]

E el 116 m

[4]

[5]

el 115 m F

Fig. P11.36 11.37

(a)

The piping system shown in Fig. P11.36 delivers water to two side-branch sprinklers (C and E) and to a lower reservoir (F). The water is supplied by an upper reservoir (A). The sprinklers are represented hydraulically as orifices with relatively high K values. Determine the flow distribution using the Hardy Cross method.

In Problem 11.36, if a pump supplying 10 kW of useful power was inserted near A, what changes in flow and pressure would occur? 11.38 A cooling water condenser flow system for a thermal power plant is shown in Fig. P11.38. Water is pumped from a reservoir at location A through a large-diameter pipe [1], passing through the condenser [2], and discharging through another largediameter pipe [3] into the receiving pond, location B. The condenser consists of a large number of elevated, small-diameter tubes placed in parallel; at either end of the condenser is a large waterfilled box called a header, C and C. Water surface elevations at A and B are known, as are the lengths and diameters of pipes 1 and 3. The condenser has N identical tubes, each with the same known diameter D2 and length L2. The friction factor for all pipes is the same constant value, and the pump curve is approximated by the relation HP a0 a1Q a2Q2 a3Q3. Minor losses can be assumed negligible.

[2] C

C⬘

A B

P [1]

[3]

Fig. P11.38

11.39 Determine the flow distribution in the branching or “tree-like” network shown in Fig. P11.39. Water is the flowing liquid: g 62.4 lb/ft3. The source of flow is a large pipeline (location A) maintained at a constant pressure of 60 psi. Each branch terminates at a location where the hydraulic grade line is known. (Courtesy of D. Wood) 11.40 The 12-pipe system shown in Fig. P11.40 represents a low-pressure region that is connected to a high-pressure system by pressure regulators set at 50 psi. Determine the pressure and water flow distribution for the demands shown. Assume Hazen–Williams C 120 for all pipes. (Courtesy of D. Wood.)


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Problems 593 15 psi 30 –– Hazen–Williams C = 110 K = 10 20 psi 40 –– 100; 2

(Length in ft; diameter in in.) 150; 2

200; 4

A

K=1

50 ––

20 80; 4 ––

200; 4

30 psi

50 ––

40 ––

200; 2

300; 2

K=2

(Elev. in ft) 60 psi

K=5

100 –––

K=3

10 ––

10 psi

Fig. P11.39

el 160 ––– 1500; 10

2 ft3/sec

150 –––

1450; 10

150 –––

1 ft3/sec 550; 10 150 ––– (Length in ft; diameter in inches)

145 –––

145 –––

1000; 10

1 150 –––

1 ft3/sec

450; 10

750; 10 Regulator (p = 50 psi)

800; 10

800; 10

155 –––

1 ft3/sec

600; 10

700; 8

Regulator (p = 50 psi)

150 –––

800; 8

400; 10

ft3/sec

140 –––

1 ft3/sec

(Elevation in ft)

Fig. P11.40


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Page 594

Chapter 11 / Flows in Piping Systems Determine the flow distribution for the 14-pipe water supply system shown in Fig. P11.41. The characteristic curve for the pump is represented by the following data (courtesy of D. Wood):

HP(m)

166

132

18

Q (L/s)

0

600

1000

Hazen–Williams roughness = 100 (all pipes) 1500; 400 P

110 L/s

K=5 610; 350

18 ––

914; 350 610; 3 50

760; 350

12 ––

K = 8 12 ––

760; 350

3 ––

30; 200 30 ––

(Elevation in m) 15 ––

914; 350

610; 350

15 ––

60 L/s

110 L/s

457;

(Diameter in mm)

0

30

300

5;

97

(Length in m)

610; 350

12 ––

34 ––

K = 10

6 ––

K = 10 610; 350

60 L/s

12 –– 60 L/s

61; 150

Fig. P11.41


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Problems 595 11.42

2. Under normal operation, pressures in the network can range between 80 and 120 psi. 3. Two gate valves are to be placed in each line (one at each end) to isolate it in case of a break or for needed maintenance. 4. All demands must be met in case of a single line break with minimum allowable pressures of 20 psi. 5. Pipes are cast iron, with the following standard diameters available: 4, 6, 8, 12, 16, 20, 24, 30, 36, 42, and 48 in. Pipes can be placed anywhere within the region.

The layout of the required water demands for a proposed industrial complex are shown in Fig. P11.42. Design an appropriate network and determine the appropriate useful power for a pump to meet the demand needs. The design criteria are as follows: 1. The lower reservoir supplies water to the system. The upper reservoir supplies water for emergency use only, such as fire, line breaks, and so on; under normal operation, there is no flow into or out of it.

Demand (gpm) Ground elevation contours (ft)

el 650 ft

80

Scale

500 ft 50 gpm

50

100

25 100

550

el 500 ft

540 P

500

75 530 520 510

Fig. P11.42


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Page 596

Chapter 11 / Flows in Piping Systems any time. As an alternative to specifying the sprinkler flow rate, specify the loss coefficient across each sprinkler (e.g., try Kv 35), and with successive trials, adjust it until the desired flow rate is achieved. 3. Water is to be pumped from the well shown. The maximum possible flow from the well is 2000 gal/min. Work with useful power in order to satisfy the system demand. 4. System pressure is to be 80 10 psi at all locations, and the design velocity in each pipe is not to exceed 5 ft/sec. Use smooth plastic piping and fittings throughout.

Design an irrigation system for part or all of the 18-hole golf course shown in Fig. P11.43. The requirements are as follows: 1. Place the sprinkler couplings at approximately 100-ft intervals along the fairway centerline. The flow rate from each sprinkler is to be 40 gal/min. A maximum of two sprinklers on each fairway are to be in operation at any time. 2. Place four sprinklers around the periphery of each green. The flow rate from each sprinkler is to be 40 gal/min. A maximum of three sprinklers on each green are to be in operation at

.

.. . .. . . .. . . . . . . . . . . .. . .. . . . .. . . . . . . . . . . . . . .. .... . ... .. ... ... . .. . . ... .. . .. ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... .... .... . . . . .. . . . . . . .... . . ..... ......... .

.....

. . . . . .850 ... . . .Mount . . . . . . . . . . . .. . . . .. . .Hope . . . . . . . . . . . . . . . . . . . . .Road . . . .. 12 .. .... 11 . 840 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. . .. 16 ... 850 . . . . . . .... . . . . . 860 .... 17 . . . . . . . . . .. . . . 10 . .. ....... . . . . . .. ....... . . . ... . . . . . . 18 ... ... .... ... . . . . . ... . . . . . . . . . ..... . ... .... ...... 3 . ..... . . . . . 870 .... 4 ..... ... ... ... ... . 15 . . . . . . . . . .. . . ... . .. 880 . . . . . ... . ..... . . . . . .. 14 ..... 5 . .. 870 Pond .. . .6.. . . .. . . . . .. 2 9 ..

....

...

...

..

...

Harrison

. . .. .... ..... ... . . . . . . . . .. . . . . . ... ..... ....... . . . . . . . . . .. ... ...... . . . .. .... . . . ..

. . . . . . . . . . . . . . . . . . .. ..

. . . . ..

. . . . . . . . . . . . .8. . . . . .

..

...

.

... .. . .

. . .....

.. . . . . . . . . . . . . . .. . . . . . .

.

. .. ............ . . . . . ........... . .. . . . . ... ... .

.. . . .. . . . . . ...

. .

.

................... .... .

860

1

..

860 Elev. contours

.. . ..

w

.. . .

. . . . . . . . . .. .

... . ... . . . . . . . . . . .. . . .... .... .... . . .. . .... . . . ... . . . ........ . . . . .. . . . . .. ...... .... . ... . . . . . . . . ... . . . .... ..... .. . .. . .. .. . ..... . . . . . ... . . . . . . . . . . ..... .... ... ... ....... . . . . . . . . .. ...... .... . . . . .. .. . .. . ... . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . .. ..... .. . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . ..

. . . . . . . ....

Fair

...... . . . . . . . ay

. . . ...

..

Green

. . . . . . . . . .. .

300 ft

... .

0

.. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...

.

7

Well pump house

. . . .. .... .... .... ..... . . . .. . ....... ......... ..... .... .. ...... . ....... ...... . . ... . . . . ... ..... ..... . .......... ....... . . ...

..

...

. ..

...

N

.. . . .

13

...

840

....... . .

11.43

11:26 AM

. . . ..

596

12/1/10

2 Hole number

Tee

Fig. P11.43


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Page 597

Problems 597

Unsteady Flow 11.44

An irrigation system has a nearly horizontal supply pipe that is connected to a tank at one end and a quick-opening valve at the other. The length of the pipe is 2500 m, and its diameter is 100 mm. If the elevation of water in the tank is 3 m, how long will it take for the flow to reach the 99% steadystate condition if the valve is suddenly opened from a closed position? Assume that the water is incompressible, the pipe is inelastic, f 0.025, and K 0.15 once the valve is opened. 11.45 In Problem 11.44 plot the velocity in the pipe as a function of time. 11.46 Gasoline is supplied by gravity without pumping from a storage tank through a 800-m-long 50-mmdiameter nearly horizontal pipe into a tanker truck. There is a quick-acting valve at the end of the pipe. The difference in elevations of gasoline between the reservoir and the truck tank is 8 m. Initially, the valve is partially closed so that K 275. Then the operator decides to increase the discharge by opening the valve quickly to the position where K 5. Assuming an incompressible fluid and an inelastic piping system, determine the new steady-state discharge and the time it takes to reach 95% of that value. Assume that f 0.015. 11.47 Water is flowing through a pipe whose diameter is 200 mm and whose length is 800 m. The pipe is made of cast iron (E 150 GPa) with a thickness of 12 mm. There is a reservoir at the upstream end of the pipe and a valve at the downstream end. Under steady-state conditions the discharge is Q 0.05 m3/s, when a valve at the end of the pipe is actuated very rapidly so that water hammer occurs. (a) How long does it take for an acoustic wave to travel from the valve to the reservoir and back to the valve?

(b) Determine the change in pressure at the valve if the valve is opened such that the discharge is doubled. (c) Determine the change in pressure at the valve if the valve is closed such that the discharge is halved. 11.48 When operating the system described in Problem 11.46, the operator decides to suddenly close the valve once steady-state flow has been reached. Assuming that the gasoline is slightly compressible and the piping is elastic, determine: (a) The acoustic wave speed. (b) The pressure rise at the valve once the valve is rapidly closed in a manner to cause water hammer. (c) If the maximum allowable pressure in the pipe is 250 kPa, what can you conclude about the outcome of the water (gasoline) hammer activity? The pipe is made of aluminum, E 70 GPa, with 2.5-mm-thick walls, and the bulk modulus of elasticity of gasoline is B 1.05 GPa. 11.49 Oil with a specific gravity of S 0.90 is flowing at 20 ft3/sec through a 20-in.-diameter 13,000-ft-long pipe. The elastic modulus of the steel pipe is E 29 106 lb/in2, its thickness is 0.40 in. and the bulk modulus of elasticity of the oil is B 217,000 lb/in2. A valve at the downstream end of the pipe is partially closed very rapidly so that a water hammer event is initiated and a pressure wave propagates upstream. If the magnitude of the wave is not to exceed 90 lb/in2, determine: (a) The percent decrease of flow rate tolerable during the valve closure. (b) The time it takes the pressure wave to reach the upstream end of the pipe.


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Grand Coulee Dam is the largest hydropower producer in the United States with a total generating capacity of 6 809 MW. It is also part of the Columbia Basin Project, irrigating more than 242 800 hectares, and is the cornerstone for water control on the Columbia River in the United States. The third phase, shown on the left, was completed in 1978; it has Francis turbines with 9.7 m runner diameters, operating at a head of 87 m and power output of 820 MW. (Courtesy of Voith Hydro)


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12 Turbomachinery Outline 12.1 Introduction 12.2 Turbopumps 12.2.1 Radial-Flow Pumps 12.2.2 Axial- and Mixed-Flow Pumps 12.2.3 Cavitation in Turbomachines 12.3 Dimensional Analysis and Similitude for Turbomachinery 12.3.1 Dimensionless Coefficients 12.3.2 Similarity Rules 12.3.3 Specific Speed 12.4 Use of Turbopumps in Piping Systems 12.4.1 Matching Pumps to System Demand 12.4.2 Pumps in Parallel and in Series 12.4.3 Multistage Pumps 12.5 Turbines 12.5.1 Reaction Turbines 12.5.2 Impulse Turbines 12.5.3 Selection and Operation of Turbines 12.6 Summary

Chapter Objectives The objectives of this chapter are to: ▲ Develop fundamental pump theory using the moment-of-momentum principle ▲ Describe radial-flow, mixed-flow, and axial-flow pumps, including presentation of prototype data and deviation from ideal conditions ▲ Introduce dimensionless pump coefficients and show how they are used in conjunction with similarity rules to develop data for turbomachinery design and analysis ▲ Demonstrate how pumps are incorporated into piping design and analysis ▲ Present introductory material on turbines: basic theory, types of turbines, and their selection and operation in conjunction with piping systems 599


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Chapter 12 / Turbomachinery

12.1 INTRODUCTION Turbomachines are the commonly employed devices that either supply or extract energy from a flowing fluid by means of rotating propellers or vanes. A turbopump, more commonly called a pump, adds energy to a system, with the result that the pressure is increased; it also causes flow to occur or it increases the rate of flow. A turbine extracts energy from a system and converts it to some other useful form, typically, to electric power. Pumps are essential components of piping systems which are designed to convey liquids. Similarly, turbomachines are called blowers, fans, or compressors when performing work on air or other gases in ducts. A hydroturbine, or simply a turbine, is a machine that generates power from high-pressure water; relatively large conduits or tunnels deliver fluid to closed turbines in order to generate power. Steam and air turbines are also of substantial engineering importance, but such devices are considered in other courses, such as thermodynamics. The examples above are all those of turbomachines designed to facilitate or utilize internal flows. A wind turbine, on the other hand, makes use of the surrounding external flow to convert the energy contained in the natural movement of atmospheric air into useful electrical power. As shown in Section 4.5.4, a propeller performs work on the surrounding fluid to provide thrust and propel an object along a desired path, while a stationary fan performs work to circulate air. All turbomachines are characterized as being capable of either adding or subtracting energy from the fluid by means of rotating propellers or vanes. There is no contained volume of fluid being transported, as in a positive-displacement piston pump. The bulk of this chapter is devoted to pumps, in particular, those that convey liquids such as water or gasoline. The centrifugal, or radial-flow pump is presented in detail, and to a lesser extent, mixed-flow and axial-flow pumps are studied. Dimensional analysis, an important tool for the selection and design of turbomachines, is presented next, followed by a discussion of the proper selection and implementation of pumps in piping systems. Finally, turbines are introduced by considering their fundamental characteristics.

12.2 TURBOPUMPS Turbopump: Pump consisting of two major parts—an impeller and the pump housing.

KEY CONCEPT Common types of pumps are radial flow, mixed flow, and axial flow.

A turbopump consists of two principal parts: an impeller, which imparts a rotary motion to the liquid, and the pump housing, or casing, which directs the liquid into the impeller region and transports it away under a higher pressure. Fig. 12.1 shows a typical single-suction radial-flow pump. The impeller is mounted on a shaft and is often driven by an electric motor. The casing includes the suction and discharge nozzles and houses the impeller assembly. The portion of the casing surrounding the impeller is termed the volute. Liquid enters through the suction nozzle to the impeller eye and travels along the shroud, developing a rotary motion due to the impeller vanes. It leaves the volute casing peripherally at a higher pressure through the discharge nozzle. Some single-suction impellers are open, with the front shroud removed. Double-suction impellers have liquid entering from both sides.


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Sec. 12.2 / Turbopumps

601

Outlet Eye Shroud

Inlet Tongue

Impeller Volute

Fig. 12.1

Casing

Single-suction pump.

In a radial-flow pump, the impeller vanes are commonly curved backward and the impeller is relatively narrow. As the impeller becomes wider, the vanes have a double curvature, becoming twisted at the suction end. Such pumps convey liquids with a lower-pressure rise than radial-flow pumps and are called mixed-flow pumps. At the opposite extreme from the radial-flow pump is the axial-flow pump; it is characterized by the flow entering and leaving the impeller region axially, parallel to the shaft axis. Typically, an axial-flow pump delivers liquid with a relatively low pressure rise. For axial-flow pumps and some mixed-flow pumps, the impellers are open; that is, there is no shroud surrounding them. Various types of impellers are shown in Fig. 12.2.

12.2.1 Radial-Flow Pumps A representative radial-flow pump is shown in Fig. 12.1; this type of pump is commonly called a centrifugal pump, and it is the most commonly used pump in existence today. An elementary analysis will be performed on a radial-flow pump, which will provide a theoretical relationship between the discharge and the developed head rise. In addition, a better understanding of the manner in which momentum exchange takes place in such a turbomachine will be provided. Actual pumps operate at efficiencies less than unity; that is, they do not perform at the theoretical, idealized conditions. It is necessary, therefore, to conduct experiments so that the true operating characteristics of a turbopump can be determined.

Axis of rotation

Radial flow

Fig.12.2

Mixed flow

KEY CONCEPT A radialflow, or centrifugal, pump is designed to deliver relatively low discharge at a high head.

Axial flow

Various types of pump impellers.


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Page 602


Elementary Theory. The actual flow patterns in a turbopump are highly three-dimensional with significant viscous effects and separation patterns taking place. To construct a simplified theory for the radial-flow pump, it is necessary to neglect viscosity and to assume idealized two-dimensional flow throughout the impeller region. Fig. 12.3a defines a control volume that encompasses the impeller region. Flow enters through the inlet control surface and exits through the outlet surface. Note that a series of vanes exists within the control volume, and that they are rotating about the axis with an angular speed v. A portion of the control volume is shown at an instant in time in Fig. 12.3b. The idealized velocity vectors are diagrammed at the inlet, location 1, and the outlet, location 2. In the velocity diagrams, V is the absolute fluid velocity, Vt is the tangential component of V, and Vn is the radial, or normal, component of V. The peripheral or circumferential speed of the blade is u vr, where r is the radius of the control surface. The angle between V and u is a. The fluid velocity measured relative to the vane is y. The relative velocity is assumed to be always tangent to the vane; that is, perfect guidance of the fluid throughout the control volume takes place. The angle between y and u is designated as b; since perfect guidance along the vane is assumed, b designates the blade angle as well.

u2

Vt2

α2 V2

ω

Vn2

β2

b2

υ2

Outlet 2

2

r

lle

ω

pe

Im

1 Inlet

b1 1 υ1

r2

V1 Vn1

β1

α1 r1

Vt1 u1 (a)

Axis of rotation (b)

Fig. 12.3 Idealized radial-flow impeller: (a) impeller control volume; (b) velocity diagrams at control surfaces.


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603

The moment-of-momentum relation, Eq. 4.6.3, can be written for steady flow in the form M

冕

c.s.

ˆ ) dA rr V(V n

(12.2.1)

Applied to the control volume of Fig. 12.3, this becomes

T rQ(r2Vt 2 r1Vt1)

(12.2.2)

in which T is the torque acting on the fluid in the control volume, and the righthand side represents the flux of angular momentum through the control volume. The power delivered to the fluid is the product of v and T: vT rQ(u2Vt2 u1Vt1)

(12.2.3)

From the velocity vector diagrams in Fig. 12.3b, Vt1 V1cos a1and Vt2 V2 cos a,2 so that Eq. 12.2.3 can be written as vT rQ(u2V2 cos a2 u1V1 cos a1)

(12.2.4)

For the idealized situation in which there are no losses, the delivered power must be equal to gQHt, in which Ht is the theoretical pressure head rise across the pump (see Eq. 4.5.26). Then there results Euler’s turbomachine relation, vT H t gQ

u2V2 cos a2 u1V1cos a1 g

(12.2.5)

Insight on the nature of flow through an impeller region can be obtained using Eq. 12.2.5. From the law of cosines we can write v 21 u12 V 21 2u1V1 cos a1 v 22 u22 V 22 2u2V2 cos a2 These can be substituted into Eq. 12.2.5 to provide the relation V 22 V 12 (u 22 u 12) (v 22 v 21) Ht 2g 2g

(12.2.6)


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The first term on the right-hand side represents the gain in kinetic energy as the fluid passes through the impeller; the second term accounts for the increase in pressure across the impeller. This can be seen by applying the energy equation across the impeller and solving for Ht: p2 p1 V 22 V 21 Ht z2 z1 g 2g

(12.2.7)

Eliminating Ht between Eqs. 12.2.6 and 12.2.7 yields the expression p1 v21 u 21 p2 v22 u 22 z1 z2 const g g 2g 2g

(12.2.8)

This relation has historically been called the Bernoulli equation in rotating coordinates. Since z2 z1 is usually much smaller than (p2 p1)/g, it can be eliminated, and thus the pressure difference is r p2 p1 (v 21 u 21) (v 22 u 22) 2

冤

冥

(12.2.9)

Returning to Eq. 12.2.5, we see that a “best design” for a pump would be one in which the angular momentum entering the impeller is zero, so that maximum pressure rise can take place. Then in Fig. 12.3b, a1 90°, Vn1 V1, and Eq. 12.2.5 becomes u2V2 cos a2 Ht g

(12.2.10)

From the triangle geometry of Fig. 12.3, V2 cos a2 u2 Vn2 cot b2, so that Eq. 12.2.10 takes the form u 22 u2 Vn 2 cot b 2 Ht g g

(12.2.11)


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605

Applying the continuity principle at the outlet region to the control volume provides the relation Q Vn2 2pr2b2

(12.2.12)

in which b2 is the width of the impeller at location 2. Introducing Eq. 12.2.12 into Eq. 12.2.11, and recalling that u2 vr2, one has the relation v2r 22 v cot b Ht 2 Q g 2pb2 g

(12.2.13)

For a pump running at constant speed, Eq. 12.2.13 takes the form Ht a0 a1Q

(12.2.14)

in which a0 and a1 are constants. Equation 12.2.13 is the theoretical head curve and is seen to be a straight line with a slope of a1, as shown in Fig. 12.4a. The effect of the blade angle b2 is shown in Fig. 12.4b. A forward curving blade (b2 90°) can be unstable and cause pump surge, where the pump oscillates in an attempt to establish an operating point. Backward-curving vanes (b2 90°) are generally preferred.

Ht

Ht Forward curving

a0

β 2 > 90° β 2 = 90°

a1 Backward curving

1

0

β 2 < 90°

Q (a)

Fig. 12.4

Q (b)

Ideal pump performance curves.


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Example 12.1 A radial-flow pump has the following dimensions: b1 44°

r1 21 mm

b1 11 mm

b2 30°

r2 66 mm

b2 5 mm

For a rotational speed of 2500 rev/min, assuming ideal conditions (frictionless flow, negligible vane thickness, perfect guidance), with a1 90° (no prerotation), determine (a) the discharge, theoretical head, required power, and pressure rise across the impeller, and (b) the theoretical head-discharge curve. Use water as the fluid. Solution (a) Construct the velocity diagram at location 1, as shown in Fig. E12.1a. The rotational speed is converted to the appropriate units as 2p v 2500 261.8 rad앛s 60 The impeller speed at r1 is then u1 vr1 261.8 0.021 5.50 m앛s u2 = 17.3 m/s υ1

V t2

V1 = 5.31m/s

β 1 = 44°

u 2 – V t2

α2

α 1 = 90° u1 = 5.50 m/s

V n2 = 3.72 m/s

β 2 = 30° υ2

V2

(a) (b) 30

H t = 19.1 m Q = 0.0077 m 3 /s

20 H t (m) 10

0

0

0.01 Q (m3/s) (c)

0.02

Fig. E12.1 From the velocity diagram we see that V1 u1 tan b1 5.50 tan 44° 5.31 m/s and since a1 90°, V1 Vn1, or Vn1 5.31 m/s. The discharge is computed to be Q 2pr1b1Vn1 2p 0.021 0.011 5.31 7.71 103 m3/s


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607

The normal component of velocity at location 2 is 7.71 103 Q Vn2 3.72 m/s 2p 0.066 0.005 2pr2b2 and the impeller speed at the outlet is u2 vr2 261.8 0.066 17.28 m/s The velocity diagram at location 2 is now sketched as shown in Fig. E12.1b. From the velocity diagram we see that Vn2 3.72 u2 Vt2 6.44 m/s tan b2 tan 30° Therefore, Vt2 u2 6.44 17.28 6.44 10.84 m/s Vn2 3.72 a2 tan1 tan1 18.9° Vt2 10.84 Vt2 10.84 V2 11.46 m/s cos a2 cos 18.9° The theoretical head is computed with Eq. 12.2.10: u2V2 cos a2 17.28 11.46 cos 18.9° Ht 19.1 m 9.81 g Hence the theoretical required power is ˙ gQH 9810 (7.71 103) 19.1 1440 W W p t The pressure rise is determined from the energy equation as follows:

冢

冣

V 21 V 22 p2 p1 Ht g 2g

冤

冥

(5.31)2 (11.46)2 19.1 9810 1.36 105 Pa 2 9.81 (b) The theoretical head-discharge curve is Eq. 12.2.13. For the present example we have (vr2)2 v cot b 2 Ht Q g 2pb2 g 261.8 cot 30° (261.8 0.066)2 Q 2p 0.005 9.81 9.81 30.4 1471Q The curve is shown in Fig. E12.1c.


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Head-Discharge Relations: Performance Curves. For real fluid flow, the theoretical head curve cannot be achieved in practice, and it is necessary to resort to experimentation to determine the actual head-discharge curve. The energy equation written across a pump from the suction side (location 1, Fig. 12.3) to the discharge side (location 2) is

冢

冣冢pg V2g z冣

p V2 HP z g 2g Ht hL

2

2

1

(12.2.15)

in which HP is the actual head across the pump, and hL represents the losses ˙ f, is through the pump. The actual power delivered to the fluid, designated as W ˙ gQH W f P Brake power: Power delivered to the impeller.

(12.2.16)

˙ , often termed the brake power, while the power delivered to the impeller is W P and is given by ˙ P vT W

(12.2.17)

˙ f would be equal to W ˙ P. Since in actuality W ˙ f W ˙ P, the If there were no losses, W 1 pump efficiency hP is defined as ˙f W gQHP hP ˙ WP vT

(12.2.18)

One objective of pump design is to make the efficiency as high as possible. The theoretical head curve is compared with the actual head curve in Fig. 12.5. The difference between the two can be attributed to the following effects: (1) prerotation of the fluid before entering the impeller region, (2) separation due to imperfect guidance of fluid as it enters the impeller region, and (3) separation due to expansion in the flow passages. Leakage and high shear rates created by the differences in velocity between fluid particles in the volute and impeller contribute further to the losses. Finally, impeller blades are designed to be most efficient at the so-called design discharge; at any other discharge—that is, “off-design”—the performance will deteriorate. 1

More precisely, hP defined by Eq. 12.2.18 is termed the overall efficiency. It is possible to define three additional efficiencies for pumps as follows: HP Hydraulic efficiency: hH Ht Q Volumetric efficiency: hV , Q QL

QL leakage flow rate

gHt(Q QL) Mechanical efficiency: hM Tv These are related to the overall efficiency by hP hH hV hM.


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609

Pump head Ht HP

Q

Fig. 12.5 Comparison between theoretical and actual radial-flow pump performance curves.

A representative pump and its performance curve are shown in Fig. 12.6; this is the curve supplied by the pump manufacturer. Typically, more than one impeller diameter will be available; in the diagram, four different impellers are represented. In addition to the head-discharge performance curves, isoefficiency, power, and net positive suction head (NPSH) curves are usually provided. Net positive suction head is associated with concern for cavitation; it is discussed in Section 12.2.3.

12.2.2 Axial- and Mixed-Flow Pumps The velocity diagram for an axial-flow pump is shown in Fig. 12.7. In the axial-flow pump, there is no radial flow and the liquid particles leave the impeller at the same radius at which they enter, so that u1 u2 u. Furthermore, assuming a uniform flow, continuity considerations give Vn1 Vn2 Vn. Equation 12.2.5, which is valid for an axial-flow pump as well as a radial-flow pump, can be combined with the identities V2 cos a2 u Vn cot b2 and V1 cos a1 Vn cot a1 to produce u2 uVn Ht (cot a1 cot b2) g g

KEY CONCEPT An axialflow pump produces relatively large discharges at low heads.

(12.2.19)

This form of the turbomachine relation is useful when the ideal absolute velocity entrance angle a1 is established by a fixed vane, or stator. If there is no prerotation, a1 90° and the theoretical head relation, Eq. 12.2.19, becomes u2 uVn cot b2 Ht g g

(12.2.20)

This relation is identical to Eq. 12.2.11, which relates the theoretical head to the impeller outlet parameters for a radial flow pump. Note, however, that Eq. 12.2.11 is valid at the periphery of the impeller for the radial-flow pump, and that all fluid particles reach the maximum head at that location. By contrast,


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0

Q (gal/min) 1000

500

1500

100 260

90 80

40

50

60

70

300

75 ηηp (%)

240

250 75

70 220

200

60 205 Hp (m)

Hp (ft)

50 Outer diameter of impeller (mm)

40

150

70

100

30 20

50 10 0

0 100 260 240 •

220

Wp (kw) 50 205

0 12 10 8 NPSH (m) 6 4 2

220

205

30

260 240

20

NPSH (ft)

10 0

50

100

150

200

250

300

350

400

0

Q (m3/h)

Fig. 12.6 Radial-flow pump and performance curves for four different impellers with N 2900 rpm (v 304 rad/s). Water at 20°C is the pumped liquid. (Courtesy of Sulzer Pumps Ltd.)


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611

Control surface 1

2

Impeller

Stationary guide vanes (a)

υ1

β1 υ1 V n1 V t1

α1 V1 1

Impeller direction

α2

V2

V t2 u2

V n2

β2 υ2 2 (b)

Fig. 12.7 Idealized axial-flow impeller: (a) impeller control volume; (b) velocity diagrams at control surface.

Eq. 12.2.20 pertains only at a specified radius for the axial-flow pump, since fluid particles enter and leave the control volume at their respective radii. In this case the head varies from a minimum at the axis to a maximum at the periphery, and the total pump head is an integrated average. Axial-flow impellers are designed to maintain a constant axial velocity throughout the impeller region; this requires that the vane angles increase gradually from the periphery to the axis, as well as from the inlet to the outlet region. Fig. 12.8 on the next page shows a representative axial-flow pump, and its performance curves.


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5000

0 0

100

200

300

10000 400

600

500

15000

700

800

900

1000

Q (gal/min) (L/s)

8 10° 10 12° 12 14° 14 16° 16 Vane angle

7

20

70 65

2

60

1

73

Hp (ft)

70 65

60

10

76

70 65 60

76

65 70 73

3

limit

76

60

60 65

4

7 65 73 0 76

Hp (m)

Oper

73

ating

5

65 70

6

70

60

65 η%

0

0 6

NPSH (m)

10° 10

12° 12

14° 14

20

16° 16

4 10

NPSH (ft)

2 0

0 50 16° 16

60

40 •

Wp (kW)

50 14° 14

30

40

12° 12

20

30

10° 10

•

Wp (hp)

20 10

10 0

500

1000

1500

2000

2500

3000

3500

0

Q (m3/h)

Fig. 12.8 Axial-flow pump and performance curves for four different vane angles with N 880 rpm (v 92.2 rad/s). Propeller diameter is 500 mm. Water at 20°C is the pumped liquid. (Courtesy of Sulzer Pumps Ltd.)


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Generally, flow in an impeller region consists of two components: the circular, vortex-like motion due to the action of the vanes, and the net flow through the impeller. The vortex motion is superposed on the radial outward flow in a radial-flow pump, and on the axial flow in an axial-flow pump. The action of the mixed-flow pump lies between these two types. The previous discussions relating to radial- and axial-flow pumps pertain to mixed-flow pumps as well. A mixedflow pump and its characteristic curves are shown in Fig. 12.9 on the next page. As in the case for a radial-flow pump, the performance characteristics of axial- and mixed-flow pumps deviate from the idealized considerations. Generally, radial-flow pumps are designed to deliver relatively low discharges at high heads, and axial-flow pumps produce relatively large discharges at low heads. Mixed-flow pumps deliver heads and discharges between these two extremes. A method of selecting a pump appropriate for required design considerations is presented in Section 12.4.

613

KEY CONCEPT A mixedflow pump delivers heads and discharges between those of radial-flow and axial-flow pumps.

Example 12.2 An axial-flow pump is designed with a fixed guide vane, or stator blade, located upstream of the impeller. The stator imparts an angle a1 75° to the fluid as it enters the impeller region. The impeller has a rotational speed of 500 rpm with a blade exit angle of b2 70°. The control volume has an outer diameter of Do 300 mm and an inner diameter of Di 150 mm. Determine the theoretical head rise and power required if 150 L/s of liquid (S 0.85) is to be pumped. Solution First, the normal velocity component Vn is Q Q 0.15 Vn 2.83 m앛s (p앛4)(0.32 0.152) A (p앛4)(D o2 D 2i) The peripheral speed u of the impeller is based on an average radius: Do Di 2p 0.3 0.15 u ⯝ v 500 5.89 m앛s 4 60 4

冢冣

The theoretical head Ht is computed with Eq. 12.2.19 to be u 5.89 Ht [u Vn(cot a1 cot b2)] [5.89 2.83(cot 75° cot 70°)] 2.46 m g 9.81 Finally, for the assumed ideal conditions, the required power is ˙ gQH (9810 0.85) 0.15 2.46 3080 W W P t

or

3.1 kW

12.2.3 Cavitation in Turbomachines Cavitation refers to conditions at certain locations within the turbomachine where the local pressure drops to the vapor pressure of the liquid, and as a result, vaporfilled cavities are formed. As the cavities are transported through the turbomachine into regions of greater pressure, they will collapse rapidly, generating extremely high localized pressures. Those bubbles that collapse close to solid boundaries can weaken the solid surface, and after repeated collapsing, pitting,

Cavitation: Condition where local pressure drops to the vapor pressure of the liquid, forming vapor-filled cavities.


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Q (m3/h) 0

500

1000

1500

2000

2500

3000

3500

20 Impeller diameter (mm)

37

18

1

35

6

16

34

60 η % 65 70

1

14 HP (m)

32

5

12

75 79 80 81

82 83

84 85

10

85 84

8 6

83 82 81 80 79

4 15 10 NPSH (m) 5 0 130 120 110 WP (kW)

100 90 80 70 0

0.1

0.2

0.3

0.4

0.5 Q

0.6

0.7

0.8

0.9

1.0

(m3/s)

Fig. 12.9 Mixed-flow pump and performance curves for four different impellers and vane angles with N 970 rpm (v 102 rad/s). Water at 20°C is the pumped liquid. (Courtesy of Sulzer Pumps Ltd.)


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615

Fig. 12.10 Cavitation bubble distribution in an impeller region. (Courtesy of Sulzer Pumps Ltd.)

erosion, and fatigue of the surface can occur. Signs of cavitation in turbopumps include noise, vibration, and lowering of the head-discharge and efficiency curves. Regions most susceptible to damage in a turbomachine are those slightly beyond the low-pressure zones on the back side of impellers (see Fig. 12.10). In general, sudden changes in direction, sudden increases in area, and lack of streamlining are the culprits causing cavitation damage in turbopumps (Karassik et al., 1986). The proper design of turbopumps will have minimized the possibility for cavitation to occur. Under adverse operating conditions, however, the pressures may decrease and cavitation may occur. Two parameters are used to designate the potential for cavitation: the cavitation number and the net positive suction head. Their interrelationship and use are now discussed. Consider a pump operating in the manner shown in Fig. 12.11. Location 1 is on the liquid surface on the suction side, and location 2 is the point of minimum pressure within the pump. Writing the energy equation from location 1 to location 2 and using an absolute pressure datum results in V 22 patm p2 z hL 2g g

(12.2.21)

in which hL is the loss between location 1 and location 2, z z2 z1, and the kinetic energy at location 1 is assumed negligible. The minimum allowable pressure at location 2 is the vapor pressure pv. Substituting this into the expression, one can say that the left-hand side of Eq. 12.2.21 represents the maximum kinetic energy head possible at location 2 when cavitation is imminent. Thus the net positive suction head (NPSH) is defined as patm pv NPSH z hL g

(12.2.22)


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Location of minimum pressure in pump

Discharge pipe Q

2 z

Inlet, or suction pipe 1

Datum

Fig. 12.11 Cavitation setting for a pump.

The NPSH is also used for turbines; however, the sign of the hL term in Eq. 12.2.22 changes, and location 1 refers to the liquid surface on the discharge side of the machine. The design requirement for a pump is thus established as follows:

patm pv NPSH z hL g

(12.2.23)

The performance data supplied by turbomachinery manufacturers usually includes NPSH curves; these are developed by testing a given family in a laboratory environment. Figures 12.6, 12.8, and 12.9 show NPSH curves. The NPSH curve enables one to specify the required maximum value of z to be used for a given turbomachine; note that it is necessary to estimate hL to obtain this. The right-hand side of Eq. 12.2.22 can be divided by HP, the total head across the pump to yield s

1 (patm pv) ¢z hL g Hp

(12.2.24)

in which s is the Thoma cavitation number. This parameter is used as an alternative to the NPSH to establish design criteria for cavitation. A critical cavitation number, which signals that cavitation is imminent, is determined experimentally. Thus, for no cavitation to occur, s must be greater than the critical cavitation number. The cavitation number is in dimensionless form, which is preferred to the dimensional form of the NPSH.


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Sec. 12.3 / Dimensional Analysis and Similitude for Turbomachinery 617

Example 12.3 Determine the elevation at which the 240-mm-diameter pump of Fig. 12.6 can be situated above the water surface of the suction reservoir without experiencing cavitation. Water at 15°C is being pumped at 250 m3/h. Neglect losses in the system. Use patm 101 kPa. Solution From Fig. 12.6, at a discharge of 250 m3/h, the NPSH for the 240-mm-diameter impeller is approximately 7.4 m. For a water temperature of 15°C, pv 1666 Pa absolute, and g 9800 N/m3. Equation 12.2.22 with hL 0 is employed to compute z to be patm pv 101 000 1666 z NPSH hL 7.4 0 2.74 m g 9800 Thus the pump can be placed approximately 2.7 m above the suction reservoir water surface.

12.3 DIMENSIONAL ANALYSIS AND SIMILITUDE FOR TURBOMACHINERY The development and utilization of turbomachinery in engineering practice has benefited greatly from the application of dimensional analysis, probably to an extent more than any other area of applied fluid mechanics. It has enabled pump and turbine manufacturers to test and develop a relatively small number of turbomachines and subsequently produce a series of commercial units that cover a broad range of head and flow demands. The material developed in Chapter 6, which covers similitude and dimensional analysis, is applied to turbomachines in this section.

12.3.1 Dimensionless Coefficients The following parameters can be considered significant ones for a turbomachine: power, rotational speed, outer diameter of the impeller, discharge, pressure change across the impeller, density of the fluid, and viscosity of the fluid. These are given in Table 12.1 along with the relevant dimensions. In functional form, the relation between these variables is given by ˙ v, D, Q, p, r, m) 0 f (W,

(12.3.1)

Using v, r, and D as repeating variables, a set of convenient dimensionless groupings can be derived. They are given as follows:


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˙ W ˙ CW rv3D5

power coefficient

(12.3.2)

p CP 22 rv D

pressure coefficient

(12.3.3)

Q CQ 3 vD

flow rate coefficient

(12.3.4)

vD2r Re m

Reynolds number

(12.3.5)

It is customary to equate p to gH, where H represents either the pressure head rise in the case of a pump or the pressure head drop for a turbine. Then Eq. 12.3.3 is replaced by

gH CH 2 2 vD

TABLE 12.1

(12.3.6)

head coefficient

Turbomachinery Parameters

Parameter Power Rotational speed Outer diameter of impeller Discharge Pressure change Fluid density Fluid viscosity

Symbol

Dimensions

˙ W v D Q p r m

ML2앛T 3 T1 L L3앛T M앛LT 2 M앛L3 M앛LT

Another dimensionless parameter for a pump can be found by grouping CQ, ˙ , and CH to form the ratio CW gQH CQCH ˙ hP ˙ CW W

KEY CONCEPT The discharge, head, and power coefficients are used to form the dimensionless performance curves.

(pump)

(12.3.7)

Hence the efficiency is also a similitude parameter. For a turbine the efficiency grouping is given by ˙ ˙ CW W hT CQCH gQH

(turbine)

(12.3.8)


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˙ is the output power.2 in which W The interrelationships between the dimensionless parameters are expressed in the form of dimensionless performance curves, as shown in Fig. 12.123; these are for the same turbomachine as shown in Fig. 12.6. Dimensionless performance curves for the axial-flow pump of Fig. 12.8 and the mixed-flow pump of Fig. 12.9 are shown on the next page in Figs. 12.13 and 12.14, respectively. The vertical dashed lines designate the operating conditions at maximum or so-called “best” efficiency. 1.5

CH 1.0 CH × 10 CNPSH × 10 CW• × 100 ηP

ηP 0.5 C W• C NPSH 0.0

0.5

1.0

1.5

2.0

2.5

CQ × 100

Fig. 12.12 Dimensionless radial-flow pump performance curves for the pump presented in Fig. 12.6; D 240 mm; P 0.61.

Note that the Reynolds number is not present as a parameter in Figs. 12.12 to 12.14, even though viscous effects are significant in turbomachinery flow. In addition to the Reynolds number, one could include a relative roughness ratio. It is known that viscous and roughness effects will alter pump behavior; however, because of the difficulties involved, it is usual practice not to make a correlation between the Reynolds number, roughness, and pump losses. The flow regime in pump passages is usually very turbulent, and the roughness varies widely for different pumps. Furthermore, the friction losses may not be as significant as the losses due to eddy formation and separation caused by diffused flow occurring in the impeller and casing. As a result, the Re effect is usually not directly represented in similitude studies. Instead, the two following performance characteristics 2 It is common industrial practice to use so-called homologous units in place of the dimensionless coefficients. The relationships between the two groupings are as follows:

Dimensionless coefficient Homologous unit

CQ

CH

˙ CW

Q 3 ND

H 22 ND

˙ W 35 N D

In the homologous units, density and gravitational acceleration have been eliminated, and the rotational speed is given in revolutions per minute, designated by N. Note that the homologous units are dimensional. 3 In the caption, P is the specific speed, which will be defined in Section 12.3.3.


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are recognized: (1) larger pumps exhibit relatively smaller losses and higher efficiencies than smaller pumps of the same family since they have smaller roughness and clearance ratios; and (2) higher viscosity liquids will cause reductions in pump head and efficiency and an increase in power requirement for a given discharge. A dimensionless net positive suction head coefficient, CNPSH, is also presented in Figs. 12.12 to 12.14. It is formed by replacing H in Eq. 12.3.6 with the NPSH defined by Eq. 12.2.22. 1.0

0.8

CH × 10 CNPSH × 10 CW• × 100 ηP

ηP

0.6

0.4 CH C NPSH 0.2 C W• 0.0 1

2

4

3

5

6

CQ × 100

Fig. 12.13 Dimensionless axial-flow pump performance curve for the pump presented in Fig. 12.8; D 500 mm; vane angle 14°; P 4.5.

1.5

CH 1.0 CH × 10 CNPSH × 10 CW• × 100 ηP

ηP 0.5 C W• C NPSH 0.0

0.5

1.0

1.5

2.0

2.5

CQ × 100

Fig. 12.14 Dimensionless mixed-flow pump performance curve for the pump presented in Fig. 12.9; D 371 mm; P 3.1.


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Example 12.4 Determine the speed, size, and required power for the axial-flow pump of Fig. 12.13 to deliver 0.65 m3/s of water at a head of 2.5 m. Solution The pump data are obtained from Fig. 12.13. Reading from the figure we find that at the design, or maximum, efficiency of hP 0.75 CQ 0.048

CH 0.018

˙ 0.0012 CW

Equations 12.3.4 and 12.3.6 are employed to determine the two unknowns v and D. Rearrange Eq. 12.3.6 to solve for vD: vD

9.81 2.5 冪莦 36.9 m앛s 冪莦 C 0.01莦 8 gHP H

Equation 12.3.4 is rearranged to include vD as a known quantity to provide D: D

0.65 莦 0.61 m 冪莦 C vD 冪莦 0.048 36.9 Q

Q

Substituting D 0.61 m into the relation vD 36.9, one finds that 36.9 36.9 v 60.5 rad앛s D 0.61

or

578 rpm

The density for water is r 1000 kg/m3. Then the pump power is determined by using Eq. 12.3.2: ˙ C ˙ rv3D5 0.0012 1000 60.53 0.615 2.2 104 W W P W Thus the speed, size, and required power are approximately 580 rpm, 610 mm, and 22 kW, respectively.

12.3.2 Similarity Rules Following the principles outlined in Chapter 6 for similitude, similarity relationships between any two pumps from the same geometric family can be developed. They are:

˙ r2 v2 3 D2 5 W (CW˙ )1 (CW˙ )2 or ˙ 2 r1 v1 D1 W1 2 H2 g1 v2 D2 2 (CH)1 (CH)2 or H1 g2 v1 D1

冢冣冢冣

冢冣冢冣

Q2 v2 D2 (CQ)1 (CQ)2 or Q1 v1 D1

(12.3.9) (12.3.10)

3

冢冣

(12.3.11)


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KEY CONCEPT Similarity rules relate variables from a family of geometrically similar turbomachines, or are used to examine changes in variables for a given machine.

These equations, called the turbomachinery similarity rules, are used to design or select a turbomachine from a family of geometrically similar units. Another use of them is to examine the effects of changing speed, fluid, or size on a given unit. They could also be used to design a pump to deliver flow on the moon or on a space station. In light of Eq. 12.3.7, one should also expect that h1 h2, but as stated previously, larger pumps are more efficient than smaller ones of the same geometric family. An acceptable empirical correlation (Stepanoff, 1957) relating efficiencies to size is

1 (hP)2 D1 1 (hP)1 D2

1앛4

冢冣

(12.3.12)

This relation can also be used for turbines by replacing hP by hT.

Example 12.5 Determine the performance curve for the mixed-flow pump whose characteristic curves are shown in Fig. 12.14. The required discharge is 2.5 m3/s with the pump operating at a speed of 600 rpm. At design efficiency, what are the power and the NPSH requirements? Water is being pumped. Solution From Fig. 12.14 the design efficiency is hP 0.85 and the corresponding design values for the coefficients are CQ 0.15

CH 0.067

C˙W 0.0115

CNPSH 0.035

The rotational speed in radians per second is p v 600 62.8 rad앛s 30 The pump diameter is computed with Eq. 12.3.4:

冢冣

Q D vCQ

冢

1앛3

2.5 62.8 0.15

冣

1앛3

0.64 m

With Eqs. 12.3.2 and 12.3.6, the required power and NPSH are computed: ˙ rv3D5C ˙ W P

W

1000 62.83 0.645 0.0115 3.1 105 W v2D2 NPSH CNPSH g 62.82 0.642 0.035 5.8 m 9.81


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Hence, the required power is 310 kW and the required NPSH is 5.8 m. The performance curve is constructed using Eqs. 12.3.10 and 12.3.11: v 22D 22 H2 (CH)1 g 62.82 0.642 (CH)1 165(CH)1 9.81 Q2 v 2D 32(CQ)1 62.8 0.643(CQ)1 16.5(CQ)1 Values of (CH)1 and (CQ)1 are read from Fig. 12.14, along with hP and placed in the table shown below. Columns 4 and 5 are Q2 and H2 computed from the similarity equations. At best efficiency, H2 10.4 m and Q2 2.54 m3/s. The characteristic curve and efficiency curve are plotted in Fig. E12.5. (CQ)1

(CH)1

0 0.019 0.039 0.058 0.077 0.096 0.116 0.135 0.154 0.174

0.138 0.123 0.110 0.099 0.091 0.088 0.085 0.076 0.063 0.046

hP

Q2(m3/s)

H2(m)

0.59 0.70 0.78 0.84 0.85 0.79

0 0.31 0.64 0.96 1.27 1.58 1.91 2.23 2.54 2.87

22.8 20.3 18.2 16.3 15.0 14.5 14.0 12.5 10.4 7.6

ηP

0.8 ηP

20

15

0.6

H2

H2 (m) 0.4

10

5

0

0

1

2

3

Q2 (m3/s)

Fig. E12.5


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12.3.3 Specific Speed

Specific speed: Dimensionless number that characterizes a turbomachine at maximum efficiency.

It is possible to correlate a turbomachine of a given family to a dimensionless number that characterizes its operation at optimum conditions. Such a number is termed the specific speed, and it is determined in the following manner. The specific speed P of a pump is a dimensionless parameter associated with its operation at maximum efficiency, with known v, Q, and HP. It is obtained by eliminating D between Eqs. 12.3.4 and 12.3.6 and expressing the rotational speed v to the first power: CQ1앛2 vQ1앛2

P 3 CH앛4 (gHP)3앛4

(12.3.13)

In Eq. 12.3.13, v is usually based on motor requirements, and the values of Q and HP are those at maximum efficiency. A preliminary pump selection can be based on the specific speed. Fig. 12.15 shows how the maximum efficiency and the discharge vary with P for radial-flow pumps. The specific speed can be correlated with the type of impeller shown in Fig. 12.2; it is given as follows:

P 1

radial-flow pump

1 P 4

mixed-flow pump

P 4

axial-flow pump

(12.3.14)

For a turbine, the specific speed T is a dimensionless parameter associated with a given family of turbines operating at maximum efficiency, with known v, ˙ T. The diameter D is eliminated in Eqs. 12.3.2 and 12.3.6, and v is HT, and W expressed to the first power to obtain

˙ T 앛r)1앛2 C1앛2 v(W ˙ W

T –––––– –––––––––– CH5앛4 (gHT)5앛4

(12.3.15)

A comparison of T for different turbines will be made in Section 12.5.4

4

In place of Eqs. 12.3.13 and 12.3.15, it has been common industrial practice to use specific speeds in dimensional form. Hence, for a pump, the dimensional, or homologous specific speed is (Ns)P NQ1/2/H3/4, and for a turbine ˙ 1/2/H5/4. For these expressions, Q is in gal/min, ft3/sec, the corresponding homologous specific speed is (Ns)T N W ˙ is in horsepower or kW, and N is in rpm. or m3/s, H is in ft or m, W


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1.0

η∞ 10,000 gal/min (2300 m3/h) 3,000 (680) 1,000 (230) (230) 1,000 500 (115) 300 (70) 200 (45) 100 (23)

0.8

50 (11)

0.6

30 (7)

ηP 0.4

0.2

0.0

0.2

0.4

0.6 Ω

0.8

1.0

P

Fig. 12.15 Maximum efficiency as a function of specific speed and discharge for radial-flow pumps. (PUMP HANDBOOK, 2ND EDITION (DURO) by Karassik. Copyright 1986 by McGraw-Hill Companies, Inc. -Books. Reproduced with permission of McGraw-Hill Companies, Inc. -Books in the format Textbook via Copyright Clearance Center.)

Another measure of cavitation is suction specific speed, S. Analogous to the formulation of specific speed of a pump, Eq. 12.3.13, suction specific speed for either a pump or turbine is given by vQ1/2 S (gNPSH)3/4

(12.3.16)

Here, two geometrically similar units will have the same S when operating at the same flow coefficient CQ. Conversely, equal values of S indicate similar cavitation characteristics when the units are operating differently. Design values of S are determined by experiment; when cavitation is not present, Eq. 12.3.16 is no longer valid.


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Example 12.6 Select a pump to deliver 500 gal/min of water with a pressure rise of 65 psi. Assume a rotational speed not to exceed 3600 rpm. Solution To estimate the specific speed we need the following: p v 3600 377 rad앛sec 30 p 65 144 HP 150 ft rg 1.94 32.2 500 Q 1.11 ft3앛sec 7.48 60 Use Eq. 12.3.13 to find the specific speed to be 苶 v兹Q

P (gHP)3앛4 377 兹1苶 .11 0.69 (32.2 150)3앛4 From Eq. 12.3.14, this would indicate a radial-flow pump. The pump of Fig. 12.12 could be used, even though the specific speed (0.61) would result in a lower efficiency. With

P ⯝ 0.61 (Fig. 12.12), the speed is estimated with Eq. 12.3.13 as

P(gHP)3앛4 v 兹Q 苶 0.61 (32.2 150)3앛4 335 rad앛sec 兹1.11 苶 Hence the required speed is 335 30/p 3200 rpm, which does not exceed 3600 rpm. The required diameter is determined with use of Fig. 12.12, where at maximum efficiency, CQ 0.0165. This is substituted into Eq. 12.3.4 to determine the diameter:

冢

Q D CQv

冢

冣

1앛3

1.11 0.0165 335

冣

1앛3

0.59 ft

12.4 USE OF TURBOPUMPS IN PIPING SYSTEMS The appropriate selection of one or more pumps to meet the flow demands of a piping system requires, in addition to a fundamental understanding of turbopumps, a hydraulic analysis of the pumps integrated into the piping system. A single pipe and pump arrangement is analyzed in Section 7.6.7. In Example 7.18 a system demand curve is obtained, and a trial solution to find the discharge and required pump head is shown. The technique is extended to simple piping


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Sec. 12.4 / Use of Turbopumps in Piping Systems 627

systems in Section 11.3, wherein it is shown that one can employ either a pump performance curve or assume a constant power input to the pump. Section 11.4 deals with piping networks. In those systems, the pump performance curve can be represented by a polynomial relation (Eq. 11.4.14) and incorporated into the linearized energy equations, or alternatively, a constant pump power requirement can be employed (Eq. 11.4.15). In this section we look more closely at the manner in which pumps are selected. The procedure involves not only satisfying the required flow and pressure requirements for a piping system, but also making sure that cavitation is avoided and that the most efficient pump is selected.

12.4.1 Matching Pumps to System Demand Consider a single pipeline that contains a pump to deliver fluid between two reservoirs. The system demand curve is defined as

冢

冣

L Q2 HP (z2 z1) f K 2 D 2gA

(12.4.1)

where atmospheric pressure is assumed at each reservoir, and the upstream and downstream reservoirs are at elevations z1 and z2, respectively. Note that z2 need not be greater than z1, and that the friction factor f can vary as the discharge (i.e., Reynolds number) varies. The term K represents the minor losses in the pipe, see Section 7.6.4. Equation 12.4.1 is represented as curve (a) in Fig. 12.16. The first term on the right-hand side is the static head and the second term is the head loss due to pipe friction and minor losses. The steepness of the demand curve is

KEY CONCEPT The intersection of the system demand curve with the pump head-discharge curve specifies the operating head and discharge.

Best operating point Head or efficiency

Efficiency (ηP)

Head (HP)

(b) System demand (increased losses) (a) System demand (design)

HD

z2 – z1

Operating point

Static head

QD

Fig. 12.16

Discharge

Pump characteristic curve and system demand curve.


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dependent on the sum of the loss coefficients in the system; as the loss coefficients increase, signified by curve (b) in Fig. 12.16, the pumping head required for a given discharge is increased. Piping systems may experience short-term changes in the demand curve such as throttling of valves, and over the long term, aging of pipes may cause the loss coefficients to increase permanently. In either case the system demand curve could change from (a) to (b) as shown. For a given design discharge, Eq. 12.3.14 allows the selection of a pump based on the specific speed. Once the type of pump is determined, an appropriate size is selected from a manufacturer’s characteristic curve; a representative curve is shown in Fig. 12.16. The intersection of the characteristic curve with the desired system demand curve will provide the design head HD and design discharge QD. It is desirable to have the intersection occur at or close to the point of maximum efficiency of the pump, designated as the best operating point.

12.4.2 Pumps in Parallel and in Series In some instances, pumping installations may have a wide range of head or discharge requirements, so that a single pump may not meet the required range of demands. In these situations, pumps may be staged either in series or in parallel to provide operation in a more efficient manner. In this discussion, it is assumed that the pumps are placed at a single location with short lines connecting the separate units. Where a large variation in flow demand is required, two or more pumps are placed in a parallel configuration (Fig. 12.17). Pumps are turned on individually to meet the required flow demand; in this way operation at higher efficiency can be attained. It is not necessary to have identical pumps, but individual pumps, when running in parallel, should not be operating in undesirable zones. For parallel pumping the combined characteristic curve is generated by recognizing that the head across each pump is identical, and the total discharge through the pumping system is Q, the sum of the individual discharges through each pump. Note A

Head

B Pumps A and B combined System demand HD Pump A

Pump B

QA

QB

Q D = ΣQ

Discharge

Fig. 12.17 Characteristic curves for pumps operating in parallel.


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the existence of three operating points in Fig. 12.17, in which pump A or pump B is used separately, or in which pumps A and B are combined. Other design operating points could be obtained by throttling the flow or by changing the pump speeds. The overall efficiency of pumps in parallel is gH Q hP D ˙P W

(12.4.2)

˙ P is the sum of the individual power required by each pump. in which W For high head demands, pumps placed in series will produce a head rise greater than those of the individual pumps (Fig. 12.18). Since the discharge through each pump is identical, the characteristic curve is found by summing the head across each pump. Note that it is not necessary that the two pumps be identical. In Fig. 12.18 the system demand curve is such that pump A operating alone cannot deliver any liquid because its shutoff head is lower than the static system head. There are two operating points, either with pump B alone or with pumps A and B combined. The overall efficiency is g(HP)QD hP ˙ WP

(12.4.3)

in which HP is the sum of the individual heads across each pump.

Head

A

B

Pumps A and B combined

System demand H D = ΣH P Pump B

Pump A

QB

Q D Discharge

Fig. 12.18 Characteristic curves for pumps operating in series.


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Example 12.7 Water is pumped between two reservoirs in a pipeline with the following characteristics: D 300 mm, L 70 m, f 0.025, K 2.5. The radial-flow pump characteristic curve is approximated by the formula HP 22.9 10.7Q 111Q2 where HP is in meters and Q is in m3/s. Determine the discharge QD and pump head HD for the following situations: (a) z2 z1 15 m, one pump placed in operation; (b) z2 z1 15 m, with two identical pumps operating in parallel; and (c) the pump layout, discharge, and head for z2 z1 25 m. z2

, f,

D L,

z1

(a)

ΣΚ

P

(b)

P P

(c)

? Fig. E12.7

Solution (a) The system demand curve (Eq. 12.4.1) is developed first:

冢

冣

L Q2 HP (z2 z1) f K 2 D 2gA Q2 0.025 70 15 2.5 2 9.81(p앛4 0.32)2 0.3

冢

冣

15 85Q2


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To find the operating point, equate the pump characteristic curve to the system demand curve, 15 85QD2 22.9 10.7QD 111QD2 Reduce and solve for QD: 2 195QD 10.7QD 7.9 0

1 2 ⬖ QD [10.7 兹10.7 苶苶95 4 1苶.9 7苶] 0.23 m3앛s 2 195 Using the system demand curve, HD is computed as HD 15 85 0.232 19.5 m (b) For two pumps in parallel, the characteristic curve is

冢冣

冢冣

Q Q HP 22.9 10.7 111 2 2

2

22.9 5.35Q 27.75Q2 Equate this to the system demand curve and solve for QD: 2 15 85QD2 22.9 5.35QD 27.75QD

112.8QD2 5.35QD 7.9 0 1 2 ⬖ QD [5.35 兹5.35 苶苶12.8 4 1苶苶] 7.9 0.29 m3앛s 2 112.8 The design head is calculated to be HD 15 85 0.292 22.2 m (c) Since z2 2 z1 is greater than the single pump shutoff head (i.e., 25 > 22.9 m), it is necessary to operate with two pumps in series. The combined pump curve is HD 2(22.9 10.7Q 111Q2) 45.8 21.4Q 222Q2 The system demand curve is changed since z2 z1 25 m. It becomes HD 25 85Q2 Equating the two relations above and solving for QD and HP results in 2 2 25 85QD 45.8 21.4QD 222QD

or 307QD2 21.4QD 20.8 0 1 2 ⬖ QD [21.4 兹21.4 苶苶07 4 3苶0.8 2苶] 0.30 m3앛s 2 307 and HD 25 85 0.302 32.7 m


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Fig. 12.19 Multistage centrifugal pump. (Courtesy of Sulzer Pumps Ltd.)

12.4.3 Multistage Pumps Instead of placing several pumps in series, multistage pumps are available (Fig. 12.19). Basically, the impellers are all housed in a single casing and the outlet from one impeller stage ejects into the eye of the next. Such pumps can provide extremely high heads. For the pump shown in Fig. 12.19, the pressure head range is 1500 to 3400 m, and the discharge can vary from 4500 m3/h down to 260 m3/h. Up to eight stages can be selected and the maximum speed is about 8000 rpm. KEY CONCEPT In contrast to pumps, turbines extract useful energy from the water flowing in a piping system.

Runner: Moving component of a turbine. Reaction turbine: Turbine using both flow energy and kinetic energy. Impulse turbine: The flow energy is converted into kinetic energy through a nozzle before the liquid impacts the runner.

12.5 TURBINES In many parts of the world, where sufficient head and large flow rates are possible, hydroturbines are used to produce electrical power. In contrast to pumps, turbines extract useful energy from the water flowing in a piping system. The moving component of a turbine is called a runner, which consists of vanes or buckets that are attached to a rotating shaft. The energy available in the liquid is transferred to the shaft by means of the rotating runner, and the resulting torque transferred by the rotating shaft can drive an electric generator. Hydroturbines vary widely in size and capacity, ranging from microunits generating 5 kW to those in large hydroelectric installations that produce over 400 MW. There are two types of turbines. The reaction turbine utilizes both flow energy and kinetic energy of the liquid; energy conversion takes place in an enclosed space at pressures above atmospheric conditions. Reaction turbines can be subdivided further, according to the available head, as either Francis or propeller type. The impulse turbine requires that the flow energy in the liquid be converted into kinetic energy by means of a nozzle before the liquid impacts on the runner; the energy is in the form of a high-velocity jet at or near atmospheric pressure. Turbines can be classified according to turbine specific speed, as shown in Fig. 12.20; the Pelton wheel is a particular type of impulse turbine (see Section 12.5.2).


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Sec. 12.5 / Turbines 633

Axis of rotation ΩT

Impulse 0 – 1.0

Francis 1.0 – 3.5

Fig. 12.20

Mixed flow 3.5 – 7.0

Axial flow 7.0 – 14.0

Various types of turbine runners.

12.5.1 Reaction Turbines In reaction turbines, the flow is contained in a volute that channels the liquid into the runner (Fig. 12.21). Adjustable guide vanes (also called wicket gates) are situated upstream of the runner; their function is to control the tangential component of velocity at the runner inlet. As a result, the fluid leaves the guide vane exit and enters the runner with an acquired angular momentum. As the fluid travels through the runner region, its angular momentum is reduced and it imparts a torque on the runner, which in turn drives the shaft to produce power. The flow exits from the runner into a diffuser, called the draft tube, which acts to convert the kinetic energy remaining in the liquid into flow energy. Francis Turbine. In the Francis turbine, the incoming flow through the guide vanes is radial, with a significant tangential velocity component at the entrance to the runner vanes (Fig. 12.22). As the fluid traverses the runner the velocity takes on an axial component while the tangential component is reduced. As it leaves the runner, the fluid velocity is primarily axial with little or no tangential component. The pressure at the runner exit is below atmospheric. The theoretical torque delivered to the runner is developed by applying Eq. 12.2.1 to the control volume shown in Fig. 12.22; the same assumptions leading to the pump torque relation, Eq. 12.2.2, apply. The resulting relation is T rQ(r1Vt1 r2Vt 2)

(12.5.1)

Multiplying the torque by the angular speed v gives the power delivered to the shaft: ˙ T vT W rQ(u1V1 cos a1 u2V2 cos a2)

(12.5.2)

The fluid power input to the turbine is given by ˙ f gQHT W

(12.5.3)

in which HT is the actual head drop across the turbine. Thus the overall efficiency is given by


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Guide vane pivot

Volute

Guide vanes

Runner

(b)

Draft tube

Tail-race (a)

(c)

Fig. 12.21 Reaction turbine (Francis type): (a) schematic; (b) Cross section of Francis turbine, Xingo hydro power plant, Brazil (Courtesy of Voith Hydro); (c) Turbine runner and housing, Amlach, Austria. (Courtesy of Voith Hydro.)


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V t1

α1

u1

1 Guide vane

β1

V1

Inlet

υ1

υ n1

Guide vane 1

2 Outlet

ω

Runner

ω 2 υ2

β2

V2

α2

V n2

r1

V t2

(a) u2

r2

Axis of rotation (b)

Fig. 12.22 Idealized Francis turbine runner: (a) runner control volume; (b) velocity diagrams at control surfaces.

˙T W vT hT ˙ Wf gQHT

(12.5.4)

The action of the guide vanes can be described by considering the velocity vector diagrams in Fig. 12.22b. Assume perfect guidance of the fluid along the guide vane; then the tangential velocity at the entrance to the runner is Vt1 Vn1 cot a1

(12.5.5)

From the velocity vector diagram the tangential velocity is also given by Vt1 u1 Vn1 cot b1

(12.5.6)

The radial velocity component can be expressed in terms of the discharge Q and the width of the runner b1: Q Vn1 2pr1b1

(12.5.7)


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Equations 12.5.5 to 12.5.7 are combined to eliminate Vt1 and Vn1. There results 2pr 12b1v cot a1 cot b1 Q

(12.5.8)

For a constant angular speed, in order to maintain the appropriate runner entry angle a1, the guide vane angle is adjusted as Q changes. Under normal operation, a turbine operates under a nearly constant head HT so that the performance characteristics are viewed differently from those of a pump. For turbine operation under constant head, the important quantities are the variations of discharge, speed, and efficiency. The interrelationships among the three parameters are shown in the isoefficiency curve of Fig. 12.23. Note the drop in discharge as the speed increases at a given guide vane setting. The efficiency is reduced by the following effects: (1) frictional head losses and draft tube head losses; (2) separation due to mismatch of flow entry angle with blade angle; (3) need to attain a certain turbine speed before useful power output is achieved; and (4) mechanical losses attributed to bearings, seals, and the like. A representative dimensionless performance curve of a Francis turbine is shown in Fig. 12.24. The speed and head are kept constant, and the guide vanes are automatically adjusted as the discharge varies in order to attain peak efficiency. 2000

Guide vane setting 35

30

1600

25 60 65 70 75 80 85 90

ηT 20

1200

80

8580 75 70 65 60

Q (L/s) 15 800

10 400 5

0 0

250

500

750

1000

1250

1500

Speed (rpm)

Fig. 12.23 Isoefficiency curve for a Francis turbine: D 500 mm, HT 50 m. (Courtesy of Gilbert Gilkes and Gordon, Ltd.)


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Sec. 12.5 / Turbines 637 1.6

1.4

0.92

1.2 CQ C Q × 10

0.90

ηT 1.0

0.88

ηr

0.86 0.8

0.84

0.6 1.6

1.8

2.0

2.2 2.4 C W• ×100

2.6

2.8

3.0

Fig. 12.24 Performance curve for a prototype Francis turbine: D 1000 mm, v 37.7 rad/s,

T 1.063, CH 0.23. (Courtesy of Gilbert Gilkes and Gordon, Ltd.)

Example 12.8 A reaction turbine, whose runner radii are r1 300 mm and r2 150 mm, operates under the following conditions: Q 0.057 m3/s, v 25 rad/s, a1 30°, V1 6 m/s, a2 80°, and V2 3 m/s. Assuming ideal conditions, find the torque applied to the runner, the head on the turbine, and the fluid power. Use r 1000 kg/m3. Solution The applied torque is computed using Eq. 12.5.1: T rQ(r1Vt1 r2Vt2) rQ(r1V1 cos a1 r2V2 cos a2) 1000 0.057(0.3 6 cos 30° 0.15 3 cos 80°) 84.4 N m Under ideal conditions, the power delivered to the shaft is the same as the fluid power input to the turbine (i.e., hT 1). Thus ˙ ˙ W W f T vT 25 84.4 2110 W

or

2.11 kW

The head on the turbine is found with the use of Eq. 12.5.3: ˙ W f HT gQ 2110 3.77 m 9810 0.057


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Axial-Flow Turbine. In an axial-flow turbine the flow is parallel to the axis of rotation (Fig. 12.25). Unlike the Francis turbine, the angular momentum of the liquid remains nearly constant and the tangential component of velocity is reduced across the blade. Both fixed-blade and pivoting-blade turbines are in use; the latter type, termed a Kaplan turbine, permits the blade angle to be adjusted to accommodate changes in head. Axial-flow turbines can be installed either vertically or horizontally. They are well suited for low-head installations. Cavitation Considerations. The net positive suction head (NPSH) and the cavitation number, defined in Section 12.2.3, are applicable for turbines. In

Guide vanes

Runner (a)

(b)

(c)

Fig. 12.25 Axial-flow turbine (Kaplan type): (a) schematic; (b) Kaplan runner for the Yacyretá hydro power plant, Argentina (Courtesy of Voith Hydro); (c) Fish-friendly Kaplan turbine, Wanapum, USA. (Courtesy of Voith Hydro.)


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Eqs. 12.2.21 to 12.2.23 the sign of the loss term becomes positive. The Thoma cavitation number, defined by Eq. 12.2.24 for a pump, is given in the following form for a turbine:

s

1 patm pn a ¢z hL b g HT

(12.5.9)

Typically, location 2 is defined at the outlet of the runner, and location 1 refers to the liquid surface at the draft tube outlet (Fig. 12.26a). The cavitation number is commonly used to characterize the cavitation behavior of a turbine. Fig. 12.26b shows a representative plot of the cavitation number versus turbine efficiency, which is obtained experimentally by the turbine manufacturer. From such a plot operational procedures related to settings of the headwater and tailwater elevations can be established, and in addition, admissible levels of the cavitation number can be determined.

12.5.2 Impulse Turbines The Pelton wheel (Fig. 12.27 on the next page) is an impulse turbine that consists of three basic components: one or more stationary inlet nozzles, a runner, and a

Inlet, or penstock

Location of minimum pressure (outlet of runner) 2 Z 1

Datum (a) 1.00

ηT

0.98 0.96 0.0

0.1

0.2

0.3

0.4

σ (b)

Fig. 12.26 Cavitation considerations: (a) schematic; (b) representative cavitation number curve.


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Chapter 12 / Turbomachinery Spear rod guide Branch pipe Spear rod

Jet shroud Top cover Runner Casting Deflector plate

Inlet pipework Spear tip Nozzle Bifurcation pipe

(a)

ω

Elevation r

Buckets Plan view of jet striking one bucket

Splitter ridge

(b)

(c) Fig. 12.27 Impulse turbine (Pelton type): (a) typical arrangement of twin-jet machine (Courtesy of Gilbert Gilkes and Gordon, Ltd.); (b) jet striking a bucket; (c) Pelton runner, Sedrun, Switzerland. (Courtesy of Voith Hydro.)


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casing. The runner consists of multiple buckets mounted on a rotating wheel. The pressure head upstream of the nozzle is converted into kinetic energy contained in the water jet leaving the nozzle. As the jet impacts the rotating buckets, the kinetic energy is converted into a rotating torque. The buckets are shaped in a manner to divide the flow in half and turn its relative velocity vector in the horizontal plane nearly 180°; the ideal angle of 180° is not attainable since the exiting liquid must stay free of the trailing buckets. The moment of momentum equation, illustrated in the preceding sections, can be applied to the control volume shown in Fig. 12.28. Neglecting friction, the torque delivered to the wheel by the liquid jet is T rQr (V1 u)(1 cos b2)

(12.5.10)

in which Q is the discharge from all jets, and u rv, r is the wheel radius as shown in Fig. 12.27b. The power delivered by the fluid to the turbine runner is ˙ T rQu(V1 u)(1 cos b2) W

(12.5.11)

Usually, b2 varies between 160 and 168°. Differentiation of Eq. 12.5.11 with respect to u and setting it equal to zero shows that maximum power occurs when u V1/2. The jet velocity can be given in terms of the available head HT: V1 Cy 兹2gH 苶 T

(12.5.12)

The velocity coefficient Cy accounts for the nozzle losses; typically, 0.92 Cy 0.98. The efficiency is ˙T W hT gQHT

(12.5.13)

Substituting Eqs. 12.5.11 and 12.5.12 into this relation and rearranging produces hT 2f(Cy f)(1 cos b2)

(12.5.14)

Control volume υ1 = V1 – u

u V1

υ2 = V1

–u

V2

u

β2

Fig. 12.28

Velocity vector diagram for Pelton bucket.


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Chapter 12 / Turbomachinery 1.0 0.8

0.6

ηT 0.4

0.2

0

0.2

0.4

0.6

0.8

1.0

φ

Fig. 12.29 Speed factor versus efficiency for a laboratory-scale Pelton turbine: solid line, Eq. 12.5.14 (Cy 0.94, b2 168°); dashed line, experimental data. 600 Nozzle setting 12 500

10 90 90 400 80

8

Q (L/s) 300

6

200

4

100

2

75 70 65 60

85

88

88

85

ηT

80 75 70 65 60

0 0

500

1000

1500 2000 Speed (rpm)

2500

3000

3500

Fig. 12.30 Isoefficiency curves for a Pelton turbine: D 500 mm, HT 500 m. (Courtesy of Gilbert Gilkes and Gordon, Ltd.)


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in which the speed factor f is defined as rv f 兹2苶 gHT

(12.5.15)

Maximum efficiency occurs at f Cy /2. Fig. 12.29 shows Eq. 12.5.14 plotted for Cy 0.94 and b2 168°. In addition, data are plotted for a laboratory-scale Pelton wheel. The theoretical maximum efficiency is 0.874, while the experimental maximum efficiency is approximately 0.80. The efficiency is reduced because (1) the liquid jet does not strike the bucket with a uniform velocity, (2) losses occur when the jet strikes the bucket and splitter, (3) frictional loss is present due to the rotation of the bucket and splitter, and (4) there is frictional loss at the nozzle. The interrelationships among Q, v, and hT are shown in the isoefficiency curves of Fig. 12.30. For a constant head and nozzle setting, the discharge remains constant, since the nozzle outlet is at atmospheric pressure. A representative dimensionless performance curve for a prototype Pelton turbine is shown in Fig. 12.31. 10

1.0 0.9

ηT

0.8 η T

8

0.7

6

0.6 CQ

CQ × 1000 4

2

0

1

2 CW• × 1000

3

4

Fig. 12.31 Performance and efficiency curves for a prototype Pelton turbine: D 1000 mm, v 75.4 rad/s, T 0.135, CH 0.52. (Courtesy of Gilbert Gilkes and Gordon, Ltd.)


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Example 12.9 A Pelton turbine rotates at an angular speed of 400 rpm, developing 67.5 kW under a head of 60 m of water. The inlet pipe diameter at the base of the single nozzle is 200 mm. The operating conditions are Cy 0.97, f 0.46, and hT 0.83. Determine (a) the volumetric flow rate, (b) the diameter of the jet, (c) the wheel diameter, and (d) the pressure in the inlet pipe at the nozzle base.

Solution (a) The discharge is computed from Eq. 12.5.13 to be ˙ W 67 500 T Q 0.138 m3앛s gHThT 9810 60 0.83 (b) From Eq. 12.5.12, the velocity of the jet is 苶苶81 9.苶0 6苶 33.3 m앛s V1 Cy 兹2gH T 0.97 兹2 The area of the jet is the discharge divided by V1, or Q 0.138 A1 4.14 103 m2 V1 33.3 Hence the jet diameter D1 is D1

4 4 A 冪莦 4.14 10 冪莦莦莦 0.0726 m p p 3

1

or

73 mm

(c) Use Eq. 12.5.15 to compute the wheel diameter D to be 2f 苶 D 2r 兹2gH T v 2 0.46 兹2苶81 9.苶0 6苶 0.754 m 400 p앛30

or

754 mm

(d) The area of the inlet pipe is p A 0.202 0.0314 m2 4 The piezometric head just upstream of the nozzle is equal to HT, so that the pressure at that location is

冢

冣

Q2 p g HT 2 2gA

0.1382 9810 60 2 2 9.81 0.0314

冢

5.79 105 Pa

or

冣

approximately 580 kPa


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12.5.3 Selection and Operation of Turbines A preliminary selection of the appropriate type of turbine for a given installation is based on the specific speed. Fig. 12.20 shows how the turbine runner varies with

T. Impulse turbines normally operate most economically at heads above 300 m, but small units can be used for heads as low as 60 m. Heads up to 300 m are possible for Francis units, and propeller turbines are normally used for heads lower than 30 m. Fig. 12.32 illustrates the ranges of application for a variety of hydraulic turbines. So-called minihydro and microhydro installations have recently been subject to renewed interest, not only in developing countries, but in industrially developed countries such as the United States, where the cost of energy has increased and incentive programs have been made available to promote their use.

1000 San Giacomo sul Vomano Guang Zhou II

100 Head (m)

Ghazi Barotha Lajeado

Pelton turbines Pump turbines Francis turbines Kaplan turbines Bulb/Pit/S turbines References

1000 MW

10

1 MW

0.1 MW

10 MW

100 MW

0 1

0

10 Flow

Fig. 12.32

100

1000

10000

(m3/s)

Application ranges for hydraulic turbines. (Courtesy of Voith Hydro.)


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Microhydro has been classified as units that possess a capacity of less than 100 kW, and minihydro refers to system capacity from 100 to 1000 kW (Warnick, 1984). It is becoming common practice for manufacturers to develop a standardized turbine unit. In addition to specialized small Pelton, Francis, and propeller units, the cross-flow turbine is employed; basically, it is an impulse turbine that possesses a higher rotational speed than that of other impulse turbines. Another alternative is to use a commercially available pump and operate it in the reverse direction; in the turbine mode, the best efficiency operating point requires a larger head and flow than when the pump operates at best efficiency in the pumping mode. Reversible pump/turbine units are used in pumped/storage hydropower installations. Water is pumped from a lower reservoir to a higher storage reservoir during periods of low demand for conventional power plants. Subsequently, water is released from the upper reservoir, and the pumps are rotated in reverse to generate power during periods of high electric power demand. Fig. 12.33 shows a representative installation consisting of a Francis reversible pump/turbine unit combined with a generator/motor. Such units require special design considerations to operate efficiently in either mode. The problem of converting the flow from the pump mode to the turbine mode, or vice versa, is a very complicated transient problem that is beyond the scope of this book. Water with enormous mass moving through the entire system must be reversed in direction; a special technique must be developed to accomplish this.

Fig. 12.33 Runner and top cover for pump turbine, Waldeck, Germany. (Courtesy of Voith Hydro.)


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Example 12.10 A discharge of 2100 m3/s and a head of 113 m are available for a proposed pumpedstorage hydroelectric scheme. Reversible Francis pump/turbines are to be installed; in the turbine mode of operation, T 2.19, the rotational speed is 240 rpm, and the efficiency is 80%. Determine the power produced by each unit and the number of units required. Solution The power produced by each unit is found using the definition of specific speed, Eq. 12.3.15. Solving for the power, we have

T ˙ r W (gHT)5앛4 T v

冤

冥

2

冤

冥

2.19 1000 (9.81 113)5앛4 240 p앛30

2

3.11 108 W

From Eq. 12.5.13, the discharge in each unit is ˙ W T Q gHThT 3.11 108 351 m3앛s 9800 113 0.8 The required number of units is equal to the available discharge divided by the discharge in each unit, or 2100/351 5.98. Hence, six units are required.

12.6 SUMMARY This chapter focuses on pumps or turbines that supply or extract energy by means of rotating impellers or vanes. First we emphasized the radial-flow, or centrifugal, pump and made use of the moment-of-momentum principle to derive an idealized head-discharge relation. We then contrasted that with the actual head-discharge performance curve, one that is obtained experimentally and is required for use in pump analysis and design. Axial-flow and mixed-flow pumps were introduced, and the differences between these and the radial-flow pump were discussed. The significance of pump efficiency and net positive suction head (relating to cavitation) diagrams were shown for the proper design and selection of pumps. Dimensional analysis and similitude were applied to both pumps and turbines; the three significant dimensionless numbers that were developed are the power coefficient, flow coefficient, and head coefficient. These coefficients were combined to yield pump or turbine efficiency, or to produce the specific speed for a pump or turbine. It was shown that a family of pumps or turbines could be represented in dimensionless form by a single curve. The grouping of pumps—either singly, in parallel, or in series—to meet appropriate head and discharge requirements for piping was illustrated. Finally, a brief introduction to turbines was provided, by relating fundamental theory along with the application of reaction, axial-flow, and impulse turbines to various hydraulic design situations.


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PROBLEMS Elementary Theory 12.1

Determine the torque, power, and head input or output for each turbomachine shown in Fig. P12.1. Is a pump or turbine implied? The following data are in common: outer radius 300 mm, inner radius 150 mm, Q 0.057 m3/s, r 1000 kg/m3, and v 25 rad/s. V 2 = 6 m/s

30°

υ 2 = 5.3 m/s

45°

45° V 1 = 3 m/s

12.4

12.5

υ 1 = 5.3 m/s

(a)

(b)

V 1 = 3.66 m/s

80°

υ 2 = 8.7 m/s

30°

12.6

30° V 2 = 6.1 m/s

υ 1 = 3 m/s

(c)

(d)

V 1 = 6 m/s

υ 1 = 3 m/s

30°

specific gravity of the pumped liquid is S 0.81, and the manometer reading is H 600 mm. Determine the pressure head rise across the pump if the discharge is Q 115 L/s. The centerlines of the suction and discharge pipes are at the same elevation. A centrifugal pump with the dimensions r2 2.5 in., b2 0.40 in., b2 60°, is pumping kerosene (S 0.80) at a rate of 2 gal/sec and rotating at 2000 rpm. For “best design” entry conditions, determine the theoretical pressure head rise and fluid power. An axial-flow pump has a stator blade positioned upstream of the impeller, and it provides an angle a1 60° to the flow. The radius of the impeller tip is 285 mm, and the hub radius is 135 mm. Determine the required average blade angle at the exit of the impeller if the pump is to deliver 0.57 m3/s of water with a theoretical head rise of 2.85 m. The rotational speed is 1500 rpm. For the system shown in Fig. P12.6, water flows through the pump at a rate of 50 L/s. The allowable NPSH provided by the manufacturer at that flow is 3 m. Determine the maximum height z above the water surface at which the pump can be located to operate without cavitating. Include all losses in the suction pipe. 7m

30°

80° V 2 = 3 m/s (e)

Pipe diameter = 10 cm, f = 0.015

υ 2 = 4.33 m/s (f)

Δz

Flanged long sweep elbows (K = 0.19)

4m

Fig. P12.1

Water (20 °C)

12.2

12.3

A centrifugal water pump rotates at 800 rpm. The impeller has uniform blade widths b1 50 mm, b2 25 mm, and radii r1 40 mm, r2 125 mm. The blade angles are b1 45°, b2 30°. Assuming no angular momentum of fluid at the blade entrance, determine the ideal flow rate, pressure head rise across the impeller, and the theoretical torque and power requirements. The suction and discharge pipes of a pump are connected to a differential mercury manometer. The diameter of the suction pipe is 200 mm and the diameter of the discharge pipe is 150 mm. The

patm = 101 kPa Reentrant entrance (K = 0.8)

Fig. P12.6 12.7

During a test on a water pump, no cavitation is detected when the gage pressure in the pipe at the pump inlet is 9.9 psi, and the water temperature is 80°F. The inlet pipe diameter is 4 in., the head


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Problems 649 (b) If the pump is to produce the same head and discharge at a location where the atmospheric pressure is 12 psi, what is the necessary change in elevation of the pump relative to the inlet reservoir to avoid cavitation?

across the pump is 115 ft, and the discharge is 13 gal/sec. (a) Compute the NPSH and the cavitation number if the atmospheric pressure is 14.7 psi.

Dimensional Analysis and Similitude 12.8

Consider the 371-mm-diameter pump curve of Fig. 12.9. If the pump is run at 1200 rpm to deliver 0.8 m3/s of water, find the resulting head rise and fluid power.

12.9

In Problem 12.8, how high above the suction reservoir can the pump running at 1200 rpm be located if the water temperature is 50°C and the atmospheric pressure is 101 kPa?

12.10

Prepare dimensionless performance curves for the 205-mm-diameter radial flow pump of Fig. 12.6. Compare these with the curves shown in Fig. 12.12. Can you explain why there may be differences between the two sets of curves?

12.11

If the 220-mm-diameter pump whose characteristics are shown Fig. 12.6 is operating at 3300 rpm, what will be the head and the NPSH when the discharge is 200 m3/h?

12.12

A pump is needed to transport 150 L/s of oil (S 0.86) with an increase in head across the pump of 22 m when running at 1800 rpm. What type of pump is best suited for this operation?

12.13

A pump is required to deliver 0.17 m3/s of water with a head increase of 104 m when operating at maximum efficiency at 2000 rpm. Determine the type of pump best suited for this duty.

12.14

Refer to Fig. 12.13, the dimensionless performance curves for an axial-flow pump. It is desired to deliver water at a rate of 45 ft3/sec and a speed of 750 rpm. Determine: (a)

The available head, diameter, power, and NPSH requirements.

(b) The actual head-discharge and powerdischarge curves. 12.15 A pump running at 400 rpm discharges water at a rate of 85 L/s. The total head rise across the pump is 7.6 m, and the efficiency is 0.7. A second pump, whose linear dimensions are two-thirds of the former, runs at 400 rpm under dynamically similar conditions; it is to be placed on a space satellite where simulated gravity produces a gravitational field which is 50% that of Earth’s. What are the discharge, head rise, and power requirements for the second pump? 12.16

A pump is designed to operate under optimum conditions at 600 rpm when delivering water at 22.7 m3/min against a head of 19.5 m. What type of pump is recommended?

12.17

A pump operating under the conditions stated in Problem 12.16 has a maximum efficiency of 0.70. If the same pump is now required to deliver water at a head of 30.5 m at maximum efficiency, determine the rotational speed, the discharge, and the required power. 12.18 A manufacturer desires to double both the discharge and head on a geometrically similar pump. Determine the change in rotational speed and diameter under the new conditions. 12.19 A pump is required to move water at 2900 gal/min against a head of 200 ft. The approximate rotational speed is 1400 rpm. Using the appropriate dimensionless performance curve in the text, determine the actual speed and size of the pump to operate at peak efficiency. 12.20 Select the type, size, and speed of pump to deliver a liquid (g 8830 N/m3) at 0.66 m3/s with fluid power requirements of 200 kW. (Hint: Assume that the pump speed can vary between 1000 and 2000 rpm.)


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Use of Turbopumps 12.21

For the system shown in Fig. P12.21, recommend the appropriate machine to pump 1 m3/s of water if the impeller speed is 600 rpm.

el 30 m L = 500

m, D = 7

50 mm, f

= 0.019,

ΣK = 3

For the pump data presented in Problem 12.22, plot the dimensionless pump curves CH versus CQ, CW˙ versus CQ, and hP versus CQ. What is the specific speed of the pump? (Note: For double-entry pumps, use one-half the discharge when computing the specific speed.)

12.26

A closed-loop flow facility (Fig. P12.26) is to be used to study water flow in a hydraulics laboratory. It is constructed of 300-mm-diameter hydraulically smooth pipe. The loss coefficient at each of the four bends is 0.1, and the total length of the loop is 14 m. The design velocity is V 3 m/s. It is proposed that an axial flow pump running at about 300 rpm be used. Verify that an axial-flow pump is the correct type to employ.

el 15 m

The following performance data were obtained from a test on a 216-mm double-entry centrifugal pump moving water at a constant speed of 1350 rev/min: Q (m3앛min)

H (m)

hP

0 0.454 0.905 1.36 1.81 2.27 2.72 3.80

12.2 12.8 13.1 13.4 13.4 13.1 12.2 11.0

0 0.26 0.46 0.59 0.70 0.78 0.78 0.74

Q

Assume that f = 0.010

ω

˙ versus Q. If Plot H versus Q, hP versus Q, and W P the pump operates in a system whose demand curve is given by 5 Q2, find the discharge, head, and power required. In the demand curve, Q is given in cubic meters per minute. 12.23 An oil (S 0.85) is to be pumped through a pipe as shown in Fig. P12.23. Using the characteristic curves of Fig. 12.12 determine the required impeller diameter, speed, and input power to deliver the oil at a rate of 14.2 L/s. Would you expect the efficiency of the pump to be the value shown on the curve?

el 12.2 m

el 0 m

12.25 P

Fig. P12.21 12.22

12.24 With reference to the pump data in Problem 12.22, if the pump is run at 1200 rpm, find the discharge, head, and required power. Is the operating efficiency under these conditions the same as the pump running at 1350 rpm?

0, ΣK =

, f = 0.01

P

75 mm 1 m, D =

L=6

Fig. P12.23

15

Fig. P12.26 12.27

In Problem 12.26, assume that the only axial pump available is one from the family shown in Fig. 12.13. If the diameter of the impeller were restricted to the pipe diameter, what would be the rotational speed and resulting head rise across the pump? 12.28 The 240-mm-diameter pump represented in Fig. 12.6 is used to move water in a piping system whose demand curve is 62 270Q2, where Q is in cubic meters per second. Find the discharge, required input power, and the NPSH requirement for: (a) One pump (b) Two pumps in parallel 12.29 It is desired to pump 600 m3/h of liquid ammonia (r 607 kg/m3) at a speed of 2500 rpm and a pressure rise of 1000 kPa. (a) Determine the type of pump best suited for this duty.


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Problems 651 (b) Using the appropriate dimensionless pump characteristic curve, find the diameter and number of stages required to meet the pumping requirements. 12.30 Crude oil (S 0.86) is to be pumped through 3 miles of 18-in.-diameter cast iron pipe. The elevation rise from the upstream to the downstream end is 640 ft. If an available pump is the 240-mmdiameter radial-flow machine of Fig. 12.6, how many pumps in series are required to provide the most efficient operation? Find the required power. 12.31 A water supply piping network designed for a community receives water from a 10-in.-diameter well (Fig. P12.31). The pump is situated 150 ft below the 10-in.-diameter primary water main that delivers the flow to the network, and due to the available ground water, the hydraulic grade line on the suction side of the pump is 75 ft above the pump. The design discharge is 500 gal/min, and a pressure of 80 lb/in2 is required in the primary water main. Making use of Fig. 12.12, determine the size, speed, and number of stages of a pump to satisfy the design requirements, and the required horsepower. Assume that the pump impeller diameter is 8 in. and that the friction factor in the well delivery pipe is constant, f 0.02. Q = 500 gpm p = 80 psi

Primary water main

D = 10 in. 150 ft

(b) What is the required NPSH of a prototype pump that is four times larger and runs at 1000 rpm? pi , Vi + + Δz

P

Fig. P12.31 12.32

A model pump (Fig. P12.32) delivers 80°C water at a speed of 2400 rpm. It begins to cavitate when the inlet pressure pi 83 kPa absolute and inlet velocity Vi 6 m/s. (a) Compute the NPSH of the model pump. Neglect losses and elevation changes between the inlet section where pi is recorded and the cavitation region in the pump.

Location of cavitation in pump

Fig. P12.32 12.33

It is required to pump water at 1.25 m3/s from a lower to an upper reservoir. The water surface in the lower reservoir is at an elevation of 20 m, and the upper reservoir elevation is 23 m. The length of the 750-mm-diameter pipe is 60 m. The friction factor can be assumed constant, f 0.02, and the entrance and exit loss coefficients are 0.5 and 1.0, respectively. (a)

What type of pump is best suited for this system if the impeller speed is to be 600 rpm?

(b) Assuming a reasonable efficiency from an appropriate pump curve, estimate the power in kilowatts required to operate the pump. 12.34

Carbon tetrachloride at 20°C is pumped from a tank opened to the atmosphere through 200 m of 50-mm-diameter smooth horizontal pipe and discharges into the atmosphere at a velocity of 3 m/s. The elevation of liquid in the tank is 3 m above the pipe. A pump is situated at the pipe inlet immediately downstream of the reservoir. (a)

75 ft

Inlet section

Q

Determine the speed and size of a pump to meet the system demand.

(b) What is the available net positive suction head at the pump inlet? Neglect losses in the short suction pipe connecting the pump to the tank. Assume that the atmospheric pressure is 101 kPa. (c) Sketch the hydraulic grade line for the system. 12.35 A wind tunnel with a test section 1 m 2 m is powered by an axial fan 1.5 m in diameter. The velocity in the test section is 20 m/s, and the rotational speed of the fan is 2500 rpm. Assuming negligible losses in the tunnel and a performance curve for the fan defined by Fig. 12.13: (a) Calculate the power required to operate the fan. (b) What is the pressure rise across the fan?


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Turbines 12.36

12.37

12.38

12.39

12.40

A Francis turbine has the following dimensions: r1 4.5 m, r2 2.5 m, b1 b2 0.85 m, b1 75°, b2 100°. The angular speed is 120 rpm, and the discharge is 150 m3/s. For no separation at the entrance to the runner, determine the guide vane angle, theoretical torque, head, and power. A model Francis turbine at 1/5 full scale develops 3 kW at 360 rpm under a head of 1.8 m. Find the speed and power of the full-size turbine when operating under a head of 5.8 m. Assume that both units are operating at maximum efficiency. A model is to be built for studying the performance of a turbine having a runner diameter of 3 ft, maximum output of 2200 kW under a head of 150 ft and a speed of 240 rpm. Determine the diameter of the model runner and its speed if the corresponding model power is 9 kW and the head is 25 ft. A hydroelectric site has available HT 80 m and Q 3 m3/s. It is proposed to use a Francis turbine unit with the operating characteristics at optimum efficiency shown in Fig. 12.24. Determine the required speed and diameter of the machine. Determine the power output, the type of turbine, and the approximate speed for the installation shown in Fig. P12.40. Neglect all minor losses except those existing at the valve.

el 650 m

is eight times the jet diameter. Use the experimental data of Fig. 12.29 at maximum efficiency to determine the required flow, wheel diameter, diameter of each jet, the number of jets required, and the specific speed. 12.43 A reaction turbine installation typically includes a draft tube (Fig. 12.26a), whose function is to convert kinetic energy of the fluid exiting the runner into flow energy. Consider a turbine operating at Q 85 m3/s and HT 31.8 m. The radius of the draft tube at the outlet of the runner is 2.5 m, and the diameter at the end of the tube is 5.0 m. Losses in the draft tube can be neglected. (a) For water at 20°C, what is the pressure at the runner outlet if in Fig. P12.43, z2 z1 2.5 m? (b) What is the permissible z2 z1 if the allowable cavitation number is s 0.14? 12.44 The Ludington pumped-storage system (Fig. P12.44) on the eastern shore of Lake Michigan consists of six penstocks delivering a combined discharge of 73,530 ft3/sec in the power-producing mode of operation. Each turbine generates 427,300 hp at an efficiency of 0.85. What diameter penstock is required for the system to operate under design conditions? Classify the type of turbine employed. el 1030 ft

HGL

Upper reservoir

el 648.5 m

Surge tank (V = 0)

Reversible pump/turbine

L = 100 m, D = 850 mm, f = 0.020 L = 2000 m, D = 850 mm, f = 0.025

Turbine ηT = 0.90

Penstock (L = 1300 ft, f = 0.01, ΣK = 0.5)

el 660 ft Lake Michigan

Kv = 1.0 el 595 m

Fig. P12.40 A water turbine is required to operate at 420 rpm under a net head of 3 m, a discharge of 0.312 m3/s, and with an efficiency of 0.9. Tests on a 1/6-scale model running at 2000 rpm are to be carried out using water. For the model turbine, determine the head drop, flow rate, output power, and expected efficiency. 12.42 A Pelton wheel develops 4.5 MW under a head of 120 m at a speed of 200 rpm. The wheel diameter

Fig. P12.44 12.45

12.41

12.46

A Francis turbine of the same family as that represented in Fig. 12.24 is running at a speed of 480 rpm under a head of 9.5 m. Determine the runner diameter, discharge, and developed power if the turbine is operating at maximum efficiency. The available head from reservoir level to the four nozzles of a twin-runner Pelton wheel is 305 m. The length of the supply pipe, or penstock, is 3 km with a friction factor f 0.02 and K 2.0. The turbine develops 10.4 MW with an efficiency of 0.85.


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Problems 653 (a)

If the head across the turbine is 95% of the available head, what is the diameter of the penstock? (b) If Cv 0.98 for each nozzle, calculate the diameter of each jet. 12.47 In a projected low-head hydroelectric scheme, 282 m3/s of water is available under a head of 3.7 m. It is proposed to use Francis turbines with specific speed 2.42, for which the rotational speed is 50 rpm. Determine the number of units required and the power to be developed by each machine. Assume an efficiency of 0.9. 12.48 Repeat Problem 12.47 if propeller turbines are substituted for the Francis units. Assume a specific speed of 4.15 and an efficiency of 0.9. 12.49 A landowner has constructed an elevated reservoir as shown in Fig. P12.49. She estimates that 1200 L/min is available as continuous flow and desires to install a small Francis turbine to generate electricity for her own use. (a) What fluid power is expected to be delivered to the turbine? (b) If a turbine similar to the one whose characteristics are shown in Fig. 12.24 is to be installed, what are the speed and size required for this installation? (c) Using an appropriate empirical equation, estimate the efficiency for this installation. What power is expected to be delivered to the generator? 1

12.50

A proposed design for a hydroelectric project is based on a discharge of 0.25 m3/s through the penstock and turbine (Fig. P12.50). The friction factor can be assumed constant, f 0.015, and the minor losses are negligible. (a) Determine the power in kilowatts that can be expected from the facility, assuming that the turbine efficiency is 0.85. (b) Show that the type of machine to be installed is a Francis turbine if the desired rotational speed is 1200 rpm. (c) Determine the size, actual speed, and actual power output of the selected turbine.

el 915 m

L = 350 m, D = 300 mm, f = 0.015

Q el 892 m T

Fig. P12.50

el 70 m

L = 105 m, D = 100 mm, f = 0.02, ΣK = 2 Q T el 47 m

2

Fig. P12.49


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Scale model of a Formula One race car in a wind tunnel. Measurements are made of flow patterns, velocity distributions, and pressures on the surface of the vehicle. Such tests are employed to reduce the drag, resulting in increased speed of the prototype vehicle. (All American Racers archives)


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13 Measurements in Fluid Mechanics Outline 13.1 Introduction 13.2 Measurement of Local Flow Parameters 13.2.1 Dynamic Response and Averaging 13.2.2 Pressure 13.2.3 Velocity 13.3 Flow Rate Measurement 13.3.1 Velocity–Area Method 13.3.2 Differential Pressure Meters 13.3.3 Other Types of Flow Meters 13.4 Flow Visualization 13.4.1 Tracers 13.4.2 Index of Refraction Methods 13.5 Data Acquisition and Analysis 13.5.1 Digital Recording of Data 13.5.2 Uncertainty Analysis 13.5.3 Regression Analysis 13.6 Summary

Chapter Objectives The objectives of this chapter are to: ▲

▲ ▲

Describe how various flow parameters are measured; these include measurements taken at a local flow position—for example, pressure or velocity—as well as integrated measurements such as discharge or pressure averaged over a cross section Present techniques and equipment employed to visualize flow fields Introduce the concepts of uncertainty analysis and error propagation resulting from collecting laboratory and field data, and demonstrate the use of regression analysis in nonlinear flow measurement

655


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Chapter 13 / Measurements in Fluid Mechanics

13.1 INTRODUCTION In the engineering laboratory and in many industrial situations, the need to measure fluid properties and various flow parameters, such as pressure, velocity, and discharge, are exceedingly important. In many instances these needs are obvious. Examples include the flow rate in a pipe or irrigation channel; the contaminant or sediment load in a river; the peak pressures on the surface of a high-rise building or the flow patterns around that building; the drag on an automobile or truck traveling at high speeds; the velocity field about a commercial aircraft; the wind velocity profile above a hilly terrain; and the size and distribution of ocean or lake waves. Fluid mechanics measurements are not exclusive to the engineer. Medical diagnosticians are interested in monitoring the functions of the cardiovascular and pulmonary systems in the human body. The movement of fluids in the agricultural, petroleum, gas, chemical, beverage, and water supply and wastewater industries require large investments annually; the uncertainties present in the measurement of these flows can have a significant impact on material and monetary considerations. Many devices have been developed to measure flow parameters; obviously, each was designed to serve a specific purpose. Before undertaking the task of measurement, it is important to define clearly the need for measuring a particular parameter. Knowledge of the underlying fluid mechanics and the physical principles involved are essential to selecting an appropriate measuring instrument and conducting a successful measurement. In addition, we must recognize that there are differences in various types of measurements. Consider, for example, discharge readings resulting from a measurement of a velocity profile integrated over a cross section; the fluctuating velocity can be time-averaged at a specific location to produce a mean value. On the other hand, instantaneous measurements of velocity in a very small sampling volume may be required to observe the transient nature of an unsteady flow. The degree of sophistication in recording a parameter can range from a simple visual reading of a manometer or dial to highspeed digital sampling of voltages representing a number of variables. The purpose of this chapter is to provide the reader with a basic introduction to the concepts and techniques applied by engineers who measure flow parameters either in the laboratory or in an industrial environment. Included is a general overview of experimental methods commonly used in fluid mechanics teaching and research laboratories; the treatment is not exhaustive, but references are provided for additional information. In the following two sections, methods and instrumentation employed to measure pressure, velocity, and discharge are presented. Next, a discussion of flow visualization is given, and the final section is devoted to data acquisition and methods of analyzing data.

13.2 MEASUREMENT OF LOCAL FLOW PARAMETERS A local flow measurement means that a quantity is measured over a relatively small sample volume of the fluid. Usually, the volume is sufficiently small so that we can say that the measurement represents the magnitude of the quantity at a point in the flow field. Two significant local flow quantities are pressure and velocity; others are


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temperature, density, and viscosity. Only the first two are discussed in this section; temperature measurement is not considered in this book. Density and viscosity would be determined from temperature and pressure measurements.

13.2.1 Dynamic Response and Averaging Flow measurements may be classified according to whether the flow is steady or unsteady. If the magnitude of a physical quantity remains constant with time, this value is referred to as the steady-state value. Conversely, if the quantity changes with time, the measurement is transient, or unsteady. Transient measurement demands a more highly specialized measuring apparatus. Most measuring instruments require a certain amount of time to respond to the physical quantity sensed. In transient measurements this response time should be much less than the time for a significant change in the physical quantity to occur. In steady-state measurements the only concern is to take a reading after the measuring system has time to respond to the flow conditions. In many situations, for example, turbulent flow, it is desired to obtain a time-average measurement of a quantity, even though the local flow conditions are unsteady; examples are Reynolds stress and turbulence intensity. In this situation, a time-averaging mechanism must be included in the measurement. In addition, an instrument must be sufficiently small to measure the spatial variation of the measured parameter, such as the wavelength of a pressure disturbance. Generally, the issue of spatial and temporal resolution must be resolved by selecting instrumentation compatible with the flow scale and with the scale of any flow disturbance; see Goldstein (1996) for further explanation.

13.2.2 Pressure Fluid pressures are measured in many different ways. The type of instrument utilized depends on the levels of precision and detail required for the particular application. Virtually all pressure measurements are based either on the manometer principle or on the concept of pressure deforming a solid material such as a crystal, membrane, tube, or plate, and then converting that deformation to an electrical signal or mechanical readout. Pressure measurements are in either the static or dynamic mode. Time-dependent pressures result from unsteadiness in the flow and from pressure disturbances; the pressure disturbances are a result of either hydrodynamic or acoustic perturbations. Often, static pressure measurements are based on time-averaged mean values, since unsteadiness is always present when the flow is turbulent. Several representative pressure gages and transducers are described below. Manometer. The manometer was analyzed in Section 2.4.3. It is a simple, inexpensive instrument for measuring static pressure, and it has no moving mechanical parts. High accuracy can be attained by using an inclined tube manometer (Fig. 13.1a) or a micromanometer (Fig. 2.6). If the pressure tap is properly machined normal to the interior wall in a duct or pipe with no burrs, and the diameter is sufficiently small (usually about 1 mm diameter), the static pressure is measured accurately (Fig. 13.1b). Often the taps are placed circumferentially around the duct in the form of a piezometer ring to average out any type of flow irregularities.


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No burrs H Manometer tubing

Wall tap (a)

(b)

Fig. 13.1 Manometer used to measure pressure: (a) inclined tube manometer; (b) piezometer opening.

Bourdon Gage. An inexpensive pressure gage for reading relatively large static pressures is the Bourdon tube (Fig. 13.2). It is a curved, flattened tube, which will expand outward when subjected to internal pressure. A mechanical linkage attached to a dial gage will provide a direct reading of the pressure. When properly designed, these gages will provide good accuracy; however, they should not be used in situations where large pressure pulsations are present, since damage to the tube may result. Pressure Transducer. Pressure transducers are electromechanical devices that are designed to measure either steady or unsteady hydrodynamic or acoustic pressures. They consist of flexible membranes or piezoelectric elements that respond to pressure changes (Fig. 13.3). All of them require accompanying electronic circuitry, which adds to their costs. Since their output is an electric signal, they can readily be interfaced with data acquisition systems for automatic storage and retrieval of data. The condenser microphone (Fig. 13.3a) is commonly utilized in aeroacoustics. Basically, it is a capacitor whose capacitance varies as the membrane is deflected due to applied pressure. These transducers are quite delicate, since they measure very small pressures (100 bar) and hence are not suitable for use in liquids. The piezoelectric transducer (Fig. 13.3b), commonly used with liquids, has a crystalline material (e.g., quartz) that produces an electric field when deformed. These rugged units can be made very small (diameters down to 2.5 mm are possible) and A

Bourdon tube

Pointer

A

Section A-A Tube deflects outward as pressure increases

Linkage

High pressure

Fig. 13.2

Bourdon tube pressure gage.


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Sec. 13.2 / Measurement of Local Flow Parameters 659 Taut membrane

Membrane Piezoelectric material

Hole for pressure equalization Metal

Glass insulator

Slotted back plate

Lead wire to transducer cable (b)

(a)

Circular membrane Strain gage

Leads to bridge

(c)

Fig. 13.3 Pressure transducers: (a) condenser microphone; (b) piezoelectric pressure transducer; (c) strain-gage transducer.

can be mounted flush with the wall of a pipe. They respond very rapidly and can measure a wide range of dynamic pressures (full vacuum to 20 MPa). A strain gage transducer (Fig. 13.3c) is another type of transducer employed with liquids; it makes use of a strain gage attached to the membrane that translates its deflection into an electrical signal. It is subject to DC drift and sensitivity changes; however, it is rugged, reliable, and insensitive to vibration.

13.2.3 Velocity The fluid velocity is of primary interest in a fluid flow. By measuring the velocity we can calculate the flow rate and perhaps even piece together an image of the streamline pattern, identifying regions of separated flow, regions of stagnated flow, and other flow characteristics. There are three categories of velocity measurements: local (at a point), spatially averaged, and velocity field measurements. The discussion in this section pertains to the first two; the third is covered in Section 13.4. Local velocities are actually measured over a small region of flow. The required precision varies, depending on the desired application. For example, time-averaged velocities measured at various points in a flow cross section may be recorded so that the velocity distribution can be integrated to provide the discharge, or flow rate. On the other hand, it may be necessary to determine the history of the time-dependent turbulent velocity components at a particular location in a region of the flow.


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Instruments for velocity measurements vary considerably in their complexity and cost, depending on the type of measurement required. The desire to measure unsteady turbulent velocity components on a relatively small local scale has popularized the use of the thermal anemometer as well as the laser-doppler velocimeter. Less complicated instruments that usually measure velocity over a large spatial region are, for example, the pitot-static probe and the propeller anemometer; they are more suitable for measuring velocities that either are steady or are slowly varying with time. Particle Velocity Measurement. Neutrally buoyant particles can be seeded in the flow to provide a visual image of the flow field; two of these, hydrogen bubbles in water and particles in air, are discussed in Section 13.4. Other tracer particles used in water are neutrally buoyant plastic beads, immiscible dyed droplets, and floats placed on a liquid surface or at a desired depth. It is important to ascertain if the particle is actually tracking the fluid motion. Very small particles that are natural impurities in the flow can be observed with appropriate optical equipment to measure velocities in many flows, especially microscopic flow fields. Pitot-Static Probe. The pitot-static probe is analyzed in Section 3.4. It measures the local mean velocity by using the Bernoulli equation. Assuming frictionless, steady flow the velocity u is given by

u

(p 莦 p) 冪2r莦

(13.2.1)

T

in which pT is the total or stagnation pressure and p is the static pressure. Equation 13.2.1 is valid only if the probe does not greatly disturb the flow, and if it is aligned with the flow direction so that the velocity u is parallel to the pitot probe. A particular design, the so-called Prandtl tube, is shown in Fig. 13.4. It has the static pressure holes positioned along the horizontal tube so that the reduced pressure resulting from flow past the nose is balanced by the increased pressure due to the vertical stem. Typically, the Reynolds number based on the pitot-tube diameter should be in the neighborhood of 1000, so that viscous effects are not Total pressure

Static pressure

Static pressure holes

V

Tube aligned with flow direction

Fig. 13.4

Pitot-static probe.


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significant. If the flow is not parallel to the probe head, the measurement error is about 1% at a yaw angle of 5°, and if the yaw angle is greater than 5°, the measurement error may be substantial. When the probe is placed close to a wall, the streamlines are deflected by the interaction of the probe with the wall; errors occur when the distance between the probe axis and the wall is less than two tube diameters. In a turbulent flow, the actual pressure is less than the sensed value; generally, the results are reliable if the fluctuation intensity is less than 10%. Other probes of a similar design are the pitot or impact tube, Preston tube, static tube, and yaw meter. The pitot tube measures the total pressure, the static tube the static pressure, and the yaw meter, the direction of flow. The Preston tube is a special form of pitot tube that has been used for the measurement of wall shear stress in boundary layers over smooth walls; basically, it is a hypodermic needle that senses the mean velocity adjacent to the wall. See Bean (1971) for details related to calibration and design. Cup or Propeller Anemometer. Cup or propeller anemometers are employed to measure the velocity of either gases or liquids. They are relatively large, and as a result, the measurement is averaged spatially over a relatively large area. These anemometers are used for many applications where extreme precision is not required. One type of cup anemometer, the current meter (Fig. 13.5a), measures the velocity in water; similar devices can measure air velocity. Cup anemometers always rotate in one sense, and thus cannot provide the flow direction. On the other hand, the propeller anemometer will reverse rotation Lead for rpm

5 rotating cups

Main support

Support arm

Main support

Vane aligns meter

(a)

(b) Rotating vanes Velocity meter

Rotating propeller

Vane aligns meter

Support strut

Shroud aligned with flow (c)

Screw-shaped blade

(d)

Fig. 13.5 Anemometers: (a) current meter; (b) propeller anemometer; (c) ducted vane anemometer; (d) wind vane anemometer.


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when the flow reverses (Fig. 13.5b); however, in contrast to cup anemometers, they must be aligned with the flow direction. A ducted propeller anemometer for measuring air velocities is shown in Fig. 13.5c. Unducted propellers are also employed; an example is the wind vane (Fig. 13.5d). Propellers are also placed inside pipes and used as flow meters. For all of these devices, counters or electromechanical readouts attached to sensors provide the rotational frequency, which is correlated to the velocity by means of a calibration test. Thermal Anemometer. The thermal (hot wire) anemometer is employed to measure velocity in air. It consists of a small wire, typically made of tungsten, platinum, or platinum-iridium, which is insulated and mounted on supports (Fig. 13.6). The wire sensor is typically 2 mm in length and 5 m in diameter, although smaller wires (0.2 mm long, 1 m in diameter) have been used. When the wire is heated, the anemometer senses the changes in heat transfer as the flow speed varies; a representative wire temperature is 250°C. The most common circuitry is one that operates in the constant-temperature mode (Fig. 13.6); earlier versions employed a constant-current arrangement. As the velocity past the probe is increased, the hot wire tends to cool, thereby attempting to lower the temperature and the resistance across the wire. The circuitry then increases the current to adjust the resistance and the probe temperature to their initial values, a process that appears to be instantaneous. Either the voltage of the amplifier or the sensor current is correlated with the velocity past the probe. The output is nonlinear with respect to velocity, and the sensitivity decreases as the velocity increases; often the signal is linearized electronically. The sensitivity as a percent of reading stays nearly constant so that a very wide range of velocities can be measured. When the fluctuation intensity is less than 20%, if the flow is incompressible and if the fluid temperature and density remain constant, the hot-wire anemometer can be used to measure a velocity component. One can then calculate the mean velocity, fluctuation intensity, and turbulence spectrum. The single-wire anemometer senses only the normal component of velocity; if the wire is properly oriented, both the mean component and the fluctuations in the mean flow direction can be measured. Two-component measurements are made using a probe with wires crossed in an “” fashion, to provide the correlation u 苶苶√苶苶 needed in turbulence measurements (u and √ are the small perturbations). Three components may be measured either by rotating the -probe or adding a third wire. Feedback for constant temperature anemometer

Wire

R1

Supports

R2 –

R

Holder

Fig. 13.6

+ R3

Thermal anemometer.


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In liquids and gases that contain solid particles, the wire probe is replaced by a more rugged cylindrical film sensor. It operates on the same principle as the hot-wire probe, but it generates turbulence in its wake because of its larger size. This self-generated turbulence can limit the probe’s capabilities for measuring actual fluctuation intensities in the flow field. In addition to measuring velocities, thermal anemometers have been adapted to monitor pressure and discharge. Laser-Doppler Velocimeter. A device that can be used advantageously when it is desired not to have the probe immersed in the flow is the laser-doppler velocimeter (LDV). The most common LDV is the dual-beam type, shown in Fig. 13.7. The frequency of the scattered light is measured by the photodetector, which converts the receiving light to an electrical signal. A simplified way to understand the principle is as follows. At the intersection of the two beams a fringe pattern is formed with fringe spacing x. Light scattered from particles passing through the fringe pattern is modulated with a frequency f, which is directly proportional to the velocity component u normal to the fringes: lf x u t 2 sin(u앛2)

(13.2.2)

in which l is the wavelength of the laser beam and u is the beam angle. Detection of flow reversal is possible by using a moving fringe pattern. The ellipsoidal measurement volume is quite small, being on the order of 0.1 mm in diameter and 1.0 mm in length. A complete analysis of the phenomenon requires a detailed description of wave optics, which is beyond the scope of this book. Velocities in both liquids and gases can be measured with an LDV. Liquids normally contain sufficient impurities which act as natural seeding agents, but gases usually are artificially seeded with extremely small liquid droplets or solid particles. Typically, an LDV can measure velocities ranging from 2 mm/s up to the supersonic range. The frequency response extends to 30 kHz, which enables Oscilloscope showing particle motion

Signal processor Flow

θ Photodetector

Laser Beam splitter

Focusing lens

Fig. 13.7

Measuring region with particles that follow fluid motion

Receiving lens

Dual-beam laser-doppler velocimeter.


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turbulent velocity components to be measured. In contrast to the thermal anemometer, the LDV requires no calibration and measures flow reversal with relative ease. The initial cost of the LDV is high, and extensive electronic instrumentation is required to process the output signal: a counter-type signal processor is most commonly used. The LDV has been employed to measure supersonic flow, flow in pumps, natural free convection, steam and gas turbine flows, combusting flows, blood flows, and open-channel flows.

13.3 FLOW RATE MEASUREMENT The measurement of flow rate—alternately termed discharge or volumetric flow— is one of the most common measurements made in flowing fluids. Numerous devices have been invented or adapted for the purpose of flow metering, ranging widely in sophistication, size, and accuracy. Basically, instruments for flow rate measurements can be divided into those that employ direct or “quantity” means of measurement, and those that are indirect, that is, the so-called rate meters. Quantity meters either weigh or measure a volume of fluid over a known time increment; examples are liquid weigh tanks, reciprocating pistons, and for gases, bellows and the liquid-sealed drum.These direct devices usually are relatively large and possess poor frequency-response characteristics. However, they do provide high precision and accuracy, and as a result, are used most often as “primary” standards for the calibration of the indirect metering devices. Indirect or rate meters consist of two components: the primary part, which is in contact with the fluid, and the secondary part, which converts the reaction of the primary part to a measurable quantity. They can be classified according to a characteristic operating principle: velocity–area measurement, pressure drop correlations, hydrodynamic drag, and the like. Rate meters are relatively low in cost, take up little space, and hence are commonly found in industrial and research laboratories. Several of the more common types are discussed below.

13.3.1 Velocity–Area Method A number of the velocity measuring devices discussed in Section 13.2.3 may be inserted in a cross section of flow and used to measure the flow rate. Thus a thermal anemometer, a laser-doppler velocimeter, a pitot-static tube, or an anemometer can be inserted in a pipe or duct of known area and calibrated to measure the flow rate. In a manner that does not require calibration and for steady flow, a velocity distribution can be measured by systematically moving a velocity probe throughout the flow cross section, or by using a so-called “rake” where local velocities are measured simultaneously. Subsequently, a numerical integration of the velocity profile over the cross-sectional area will yield the discharge. This technique has been used in both closed conduit and open-channel flows.

13.3.2 Differential Pressure Meters Differential pressure meters are widely used in industrial applications and laboratories because of their simplicity, reliability, ruggedness, and low cost. Three


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Sec. 13.3 / Flow Rate Measurement 665

commonly used types are discussed in this section: the orifice meter, venturi meter, and flow nozzle. Their operation is based on the principle of an obstruction to flow being present in a duct or pipe, and consequently, a pressure differential will exist across the obstruction. This pressure drop can be correlated to the discharge by means of a calibration, and subsequently, the pressure-discharge curve can be used to determine the discharge by reading the differential pressure. In this section we deal only with the discharge of incompressible fluids in circular pipes. The fundamental discharge relation for the differential pressure meter can be described in the following manner. Fig. 13.8 represents a thin-plate orifice meter, which can be considered as a representative differential pressure meter. A steady flow occurs in a circular duct, encounters the restrictive orifice with area A0, and issues as a downstream jet. Downstream of the restriction, the streamlines converge to form a minimal flow area Ac, termed the vena contracta. Pressure taps are located at two positions: upstream of the restriction in the undisturbed flow region (location 1) and downstream at some location in the vicinity of the vena contracta (location 2). Assuming a frictionless, incompressible fluid, the Bernoulli equation applied along the center streamline from the upstream location to the vena contracta is V 21 p1 V 2c pc z1 zc 2g g 2g g

(13.3.1)

Similarly, the continuity equation is (13.3.2)

V1A1 VcAc Combining Eqs. 13.3.1 and 13.3.2, and solving for Vc yields Vc

莦冪莦 1 (A莦 앛A ) 2g(h1 hc) c

1

2

(13.3.3)

Plane of vena contracta with area Ac Upstream pressure p1

p2

Downstream pressure

Orifice plate with area A0

Fig. 13.8

Flow through an orifice meter.


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in which p1 h1 z1 g

pc hc zc g

(13.3.4)

The ideal discharge Qi is equal to the area multiplied by the average velocity at the vena contracta: Qi AcVc Ac 2 兹2g(h 苶 hc) 1 苶兹苶 1 (A苶 c 앛A1)

(13.3.5)

The actual discharge differs from the ideal for two primary reasons. Because of real fluid flow, friction causes the velocity at the centerline to be greater than the average velocity at each cross section. Second, the piezometric head hc, evaluated at the vena contracta in the relation, is substituted with h2, the known reading at the downstream pressure tap. Also, since the area of the vena contracta is unknown, it is convenient in Eq. 13.3.5 to replace Ac by CcA0, where Cc is the contraction coefficient and A0 is the area of the orifice opening. These anomalies are accounted for by introducing a discharge coefficient Cd, which is the product of the contraction coefficient and a velocity coefficient, so that the actual discharge Q is given by the relation Cd A0 Q 2 兹2g(h 苶 h2) 1 苶兹苶 1 (Cc苶 A0 앛A苶 1)

(13.3.6)

For a circular cross section, which is typical of most differential pressure meters, it is convenient to introduce the diameter ratio b

冪莦 A D A0

D0

(13.3.7)

1

where D is the pipe diameter. A convenient way to express Eq. 13.3.6 is Q KA0 兹苶 2g(h1 苶 h2)

(13.3.8)

in which K is the flow coefficient Cd K 兹苶 1 C2cb4

(13.3.9)

A dimensional analysis would reveal that Cd and K are dependent on the Reynolds number. It is convenient to evaluate the Reynolds number either at the approach region or at the obstruction.


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Orifice Meter. A thin plate orifice meter (Fig. 13.9) is typically manufactured in the range 0.2 b 0.8. In the figure, two means of locating the pressure taps are shown: (1) flange taps, positioned 25 mm upstream and downstream of the orifice plate, and (2) taps placed one diameter upstream and one-half diameter downstream of the plate. The second arrangement is preferred, since it is capable of sensing a larger differential pressure, and it conforms better to geometric similarity laws. A third arrangement, not shown in the figure, has the pressure taps located in the pipe wall immediately upstream and downstream of the orifice; taps placed at this location have been termed corner taps. Fig. 13.10 shows experimentally determined values of the flow coefficient K for orifices as a function of b and Re0. These data were obtained using corner taps; however, data taken with flange taps or with D : D/2 taps would be indistinguishable when plotted on Fig. 13.10. If greater precision is desired, numerical data for the different taps are provided in Bean (1971). Notice that for a given b, K becomes nearly constant at high Reynolds numbers, but as the Reynolds number becomes lower, K first increases to a maximum and then decreases. Maximum values of K occur at Reynolds numbers between 100 and 1000, depending on the value of b; here K is dominated by the reduced area of the vena contracta. If the discharge is known, the Reynolds number at the orifice is known. Hence, with Fig. 13.10 one can read K directly, and subsequently determine (h1 h2) from Eq. 13.3.8. However, one would more likely use the figure in conjunction with Eq. 13.3.8 to determine the flow rate, given that (h1 h2) has been read from an attached manometer or pressure transducer. In that situation, K is not known a priori, since it depends on Re0. One can initially estimate K based on an assumed Re0 (usually, assume Re0 to be large), and subsequently by trial and error improve on that estimate by successive substitution into Eq. 13.3.8. D

D/2 D and D/2 taps

D

D0

0.01 to 0.02D

Flange taps 25 mm 25 mm

Fig. 13.9 Details of a thin-plate orifice meter. (FLUID MECHANICS MEASUREMENTS by G. E. Mattingly. Copyright 1996 by Taylor & Francis Group LLC-Books. Reproduced with permission of Taylor & Francis Group LLC-Books in the format Textbook via Copyright Clearance Center.)


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1.2

Venturi meters and nozzles 1.1 D0 β = ––– D

β = 0.6 β = 0.5

1.0

β = 0.4

0.9 K Orifices β = 0.80

0.8

β = 0.40 β = 0.70 0.7

β = 0.60 β = 0.20 0.6

101

β = 0.10

102

103

β = 0.50

104

105

106

4Q Re0 = ––––– π D0v

Fig. 13.10 Flow coefficient K versus the Reynolds number for orifices, nozzles, and venturi meters. (Adapted from Engineering Fluid Mechanics, Robertson and Crowe, © 1990 John Wiley & Sons, Inc., New York. Reproduced with permission of John Wiley & Sons, Inc.)

However, if repeated readings are to be taken, it is more convenient to determine a calibration formula of the form Q C(h1 h2)m

(13.3.10)

in which C and m are constants determined by a “best fit” criterion. See Section 13.5.3 for details of one well-known criterion, the method of least squares. The orifice has become the most widely used differential meter for measuring flow rates in liquids. This is understandable, since it is relatively inexpensive


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and can easily be placed in an existing pipe. One disadvantage is that it produces a large head loss that cannot be recovered downstream of the orifice. Venturi Meter. The venturi meter has a shape that attempts to mimic the flow patterns through a streamlined obstruction in a pipe. The classical, or Herschel, type of venturi meter is rarely used today, since its dimensions are rather large, making it cumbersome to install and expensive to fabricate. It is made up of a 21° conical inlet contraction, followed by a short cylindrical throat, leading to a 7° or 8° conical exit expansion. The discharge coefficient is nearly unity. By contrast, the contemporary venturi tube, shown in Fig. 13.11, consists of a standard flow nozzle inlet section [ISA 1932 standard (Bean, 1971)] and a conical exit expansion no greater than 30°. Its recommended range of Reynolds numbers is limited from 1.5 105 to 2 106. Equation 13.3.8 is valid for the venturi as well as for the orifice; representative values of the flow coefficient K are shown in Fig. 13.10. Because of streamlining of the flow passage, the head loss in the venturi meter is much less than in the orifice. The vena contracta is not present, and as a result the discharge coefficient Cd remains close to unity. Flow Nozzle. The flow nozzle is illustrated in Fig. 13.12. It consists of a standardized shape with pressure taps typically located one diameter upstream of the inlet and one-half diameter downstream. There are two standard shapes, either the long radius or the short radius. Due to the lack of an expansion section downstream of the nozzle, the total head loss is similar to that of an orifice, except that the vena contracta is nearly eliminated and the discharge coefficient is nearly unity. Similarly, when the pressure taps are located at D:D/2, the flow coefficient K varies with the Reynolds number in a manner nearly identical to that for the venturi meter, as shown in Fig. 13.10. The flow nozzle has an advantage over the orifice plate in that it is less susceptible to erosion and wear, and relative to the venturi meter, it is less expensive and simpler to install. Elbow Meter. A relatively simple meter can be fabricated by locating pressure taps at the outside and inside of a pipe elbow (Fig. 13.13). By applying the linear momentum equation to a control volume encompassing the fluid in the elbow (or Euler’s equation normal to the streamlines), one can derive the flow equation Q KA

p1

R 冪莦 D r p

(13.3.11)

p2 15 to 30°

Pipe with diameter D

Throat with diameter D0

Pipe with diameter D

Fig. 13.11 Venturi meter.


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Chapter 13 / Measurements in Fluid Mechanics D

D/2

p1

p2

D0

D

Fig. 13.12 Flow nozzle.

in which R is the radius of curvature of the elbow and p is the pressure difference generated by centrifugal force across the bend. The flow coefficient K can be most accurately determined by in-place calibration of the meter. However, a relation for K has been reported (Bean, 1971) for 90° elbows with pressure taps located in a radial plane 45° from the inlet: 6.5 K 1 兹R 苶 e

(13.3.12)

which is valid when 104 Re VD/n 106, and R/D 1.5. The elbow meter costs much less than any of the differential pressure meters, and it usually does not add to the overall head loss of the piping system since elbows are often present in any case.

13.3.3 Other Types of Flow Meters Turbine Meter. The turbine meter consists of a propeller mounted inside a duct that is spun by the flowing fluid. The propeller’s angular speed is correlated to the discharge; the angular rotation is measured in a manner identical to the anemometer discussed in Section 13.2.3. Accuracies up to 0.25% of the flow rate are attainable over a relatively large discharge range (Goldstein, 1996). Viscous effects become a limiting factor at the low end, whereas at the high end, accuracy is limited by the interaction between the blade tips and the electronic sensors. Each type of turbine meter must be calibrated individually. They are commonly employed to monitor flow rates in fuel supply lines. po

D

Δ p = po – pi

pi

R

Fig. 13.13 Elbow meter.


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Tapered body

Float Scale Float guide

Float stop

Fig. 13.14 Rotameter.

Rotameter. The rotameter consists of a tapered tube in which the flow is directed vertically upward (Fig. 13.14). A float moves upward or downward in response to the flow rate until a position is reached where the drag force on the float is in equilibrium with its submerged weight. Calibration consists of correlating the vertical elevation of the float with the discharge. With appropriate design, the float position can be made linearly proportional to the discharge, or if desired, another relation, such as position logarithmically proportional to discharge, can be formulated. The head loss depends on the friction loss of the tube plus the loss across the floating element. The rotameter does not provide accuracy as good as the differential pressure meters; typically it is in the range of 5% full scale. Target Meter. Fig. 13.15 shows a target meter, which consists of a disk suspended on a support strut immersed in the flow. The strut is connected to a lever arm, or alternatively, has a strain gage bonded to its surface. Fluid drag on the disk will cause the strut to flex slightly, and the recorded drag force can be related to the discharge. The assembly must be sufficiently stiff so that the drag can be measured without movement or rotation of the disk; otherwise, the drag characteristics would be altered. One advantage of this device is that it is possible to record flow reversal by measuring the sense of the drag force; hence it can be used as a bidirectional flow meter. Drag meters are fairly rugged and can be used to measure discharges in sediment-laden fluids. Electromagnetic Flowmeter. This is a nonintrusive device that consists of an arrangement of magnetic coils and electrodes encircling the pipe (Fig. 13.16). The coils are insulated from the fluid and the electrodes make contact with the fluid. Sufficient electrolytes are dissolved in the fluid so that it becomes capable of conducting an electric current. When it passes through the magnetic field generated by the coils, the liquid will create an induced voltage proportional to the flow. Electromagnetic flowmeters have been used for many applications, including blood flow and seawater measurements. Commercial models are


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Chapter 13 / Measurements in Fluid Mechanics Force transducer Lever arm Drag disk

Fig. 13.15 Target meter.

available in a wide range of diameters; they are costly and can be used only with liquids. Acoustic Flowmeter. The ultrasonic, or acoustic, flowmeter is based on one of two principles. The first one uses ultrasonic transmitters/receivers beamed across a flow path (Fig. 13.17). The measured differences in the travel times (or inversely, the frequencies) are directly proportional to the average velocity, and consequently, the discharge. The second type is based on the Doppler effect, in that acoustic waves are transmitted into the flow field and subsequently scattered by seeded particles or contaminants. The Doppler shift recorded between the transmitter and the receiver is then related to the discharge. Like the electromagnetic flowmeter, acoustic flowmeters are nonintrusive and have been used in both piping and free-surface flow systems. Vortex-Shedding Meter. This flow-metering device consists of a single strut or series of struts placed chord-like inside a pipe normal to the flow direction. If the Reynolds number is above a threshold value, vortices appear in the near-wake regions behind the struts (see Section 8.3.2). The periodic vortex shedding that occurs behind the struts can be correlated to the Reynolds number as shown in Fig. 8.9. Different devices are employed to detect the vortex-shedding frequency. Examples include a pressure transducer mounted on the downstream surface of the strut, a thermal sensor that detects changes in heat flux on the strut surface, or a strain gage that senses oscillation of the strut. Vortex-shedding meters are can be specified to operate in flows that range up to two orders of magnitude. Signal processor Magnetic field coils Magnetic field Electrodes Fluid flow through pipe

Fig. 13.16 Electromagnetic flowmeter.


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Sec. 13.4 / Flow Visualization

673

Receiver/transmitter

Acoustic signal

Transmitter/receiver

Fig. 13.17

Acoustic flowmeter.

Coriolis-Acceleration Flowmeter. These are meters that are based on the Coriolis acceleration principle (see Eq. 3.2.15). Typically they are used to measure liquid flows in piping. A U-shaped tube filled with the liquid to be measured is vibrated normal to the plane of the U and with the tips of the U stationary. The liquid flowing in the tube produces a torsional motion about the axis of symmetry of the tube. The amplitude of the motion is proportional to the flow rate, and its frequency can be correlated to the density of the liquid. The meter provides both flow rate and density measurements in a noninvasive manner. Open-Channel Flowmeters. In Section 10.4.3, flow measurement in open-channel systems is presented. Discharge relations for the broad-crested weir and sharp-crested weir are developed, and additional methods of flow measurement are discussed. The reader is referred to that section for detailed information.

13.4 FLOW VISUALIZATION Transparent flow fields have been visualized in many different ways. This is well illustrated by the hundreds of photographs presented in the publication An Album of Fluid Motion (Van Dyke, 1982). The book shows flows visualized with smoke, dye, bubbles, particles, shadowgraphs, schlieren images, interferometry, and other techniques. Usually, a method is selected that best shows the flow features of interest. Examples include particles used to visualize pathlines in liquid flows around submerged objects, dye released to study the mixing process in a stream, smoke released at the upwind end of a wind tunnel to study the development of a boundary layer, and airflow patterns above a solid surface visualized by coating that surface with a viscous liquid (streaky lines are formed that coincide with the streamlines near the surface). Several of these techniques are discussed in this section. Pathlines, streaklines, and streamlines have been defined in Chapter 3. The purpose of defining them was to provide assistance in describing the motion of a flow field. A pathline is physically generated by following the motion of an individual particle such as a bubble or small neutrally buoyant sphere over a period of time; this could be achieved by time-lapse photography or video recording. On the other hand, if a trace from smoke, a train of extremely small bubbles, or dye


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continuously emanating from a stationary source is photographed or recorded, a streakline is observed. When the generation of a streakline is interrupted periodically, time streaklines are produced.

13.4.1 Tracers Tracers are fluid additives that permit the observation of flow patterns. An effective tracer does not alter the flow pattern but is transported with the flow and is readily observable. It is important that tracers are not affected by gravitational or centrifugal forces resulting from density differences. Furthermore, their size should be at least one order of magnitude smaller than the length scale of the flow field. Several of them are discussed below. Hydrogen Bubbles. A very thin metallic wire can be placed in water to serve as the cathode of a direct-current circuit, with any suitable conductive material acting as an anode. When a voltage is supplied to the circuit, hydrogen bubbles will be released at the cathode and oxygen bubbles at the anode. The primary reaction is the electrolysis of water in the manner 2H2O씮2H2 O2. Usually, the hydrogen bubbles are used as the tracer since they are smaller than the oxygen bubbles and more of them are formed. The bubbles are transported by the flow field away from the cathode wire as a continuous sheet. If the voltage is pulsed, discrete streaklines are formed; Fig. 13.18 shows such a pattern. Standard water supplies usually contain sufficient electrolytes to sustain a current, although addition of an electrolyte such as sodium sulfate greatly increases bubble generation. Use of extremely fine wire (0.025 to 0.05 mm in diameter) with adequate velocities (Reynolds numbers based on the wire diameter should be less than 20) may create bubbles sufficiently small to negate effects of buoyancy. Chemical Indicators. Certain organic chemicals change color when the pH of water changes. For example, thymol blue solution is yellow at pH 8.0 and

Fig. 13.18 Hydrogen bubble streaklines showing separated flow around a rotating airfoil. (Courtesy of M. Koochesfahani.)


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675

Fig. 13.19 Turbulent wake behind a circular cylinder at Re 1760. (Courtesy of R. E. Falco.)

blue at pH 9.2; phenol red solution is yellow at pH 6.8 and red at pH 8.2. Thus a change in pH caused by the injection of a base solution changes the color of the liquid. The lifetime of the colored water depends on the molecular diffusion of the hydrogen ions and the turbulence. An advantage of this technique is that color residues can be eliminated by changing the pH. One application is in studies of solid–liquid phase transport phenomena; if the chemical indicator is in contact with a metal surface and the metal surface is given a negative charge, the solution in immediate contact with the plate will change color. Particles in Air. Particles in air may be introduced as helium bubbles. By properly selecting the mixture of soap and water, one can produce helium-filled soap bubbles that are nonbuoyant; it is possible to produce bubble diameters on the order of 4 mm. Solid particles or liquid droplets could also be utilized, but they must be extremely small to avoid gravitational effects. With these small sizes one must employ an extremely strong light source to visualize the flow. Smoke has been used successfully to study the detailed structure of complex flow phenomena. It is the most popular agent used for flow visualization in wind tunnels. One injection technique is the so-called smoke-wire method, where the smoke is generated by vaporizing oil from a fine electrically heated wire. The method can be applied to flows where the Reynolds number based on the wire diameter is less than 20. An example of such a flow structure is given in Fig. 13.19. Smoke can also be released from a small-diameter tube or “rake” to create one or more streaklines. Velocimetry. Simultaneous measurement of two- and three-dimensional flow domains has led to the definition of a generic pulsed light velocimeter (Fig. 13.20). It consists of a pulsed light source that illuminates small fluid-entrained particles over short exposure times, and a camera that is synchronized with the light to record the location of the particles. The velocity of each marker is then given by s/t, where s is the displacement of the marker and t is the time between exposures. The technique is called Particle Image Velocimetry (PIV). The markers are usually particles that vary in size from 1 to 20 m. An example of PIV is shown in Fig. 13.21 on page 677. In addition to PIV, progress has been made in the use of molecular markers such as photochromic dyes in liquids, photochromic aerosols in gases, and molecular phosphorescence in both liquids and gases. A sample of Molecular Tagging Velocimetry (MTV) is shown in Fig. 13.22 on page 678, in which a noninvasive grid is tagged and subsequently displaced during


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Light sheet y x Flow

z

Spot of interest

Lens

Measurement volume

Y X

Image plane of camera

Fig. 13.20 Pulsed light velocimeter. (Courtesy of R. Adrian.)

a brief interval. The two grids are captured with a CCD camera and analyzed to produce the velocity field shown in Fig. 13.22c. In that figure, only the left half of the velocity field is shown. The axis of symmetry is indicated by the dashed line, and the box delimits the region containing vectors derived from Figs. 13.22a and b. Velocimetry methods are particularly useful in the studies of explosions, transients in channels and ducts, flow in internal combustion engines, bubble growth and collapse, fluid-solid interaction, and the structure of turbulent flow. Tufts and Oil Films. Flow visualization along a surface can be accomplished by attaching tufts to the surface, or if flow away from the surface is to be observed, they may be supported on wires. Surface tufts may be used to observe the transition from laminar to turbulent motion. They may also be used to study flow separation qualitatively, where the violent motion of the tufts or their tendency to point in the upstream direction identifies that separation is taking place. In air, surface flow patterns can also be visualized by distributing oil over the solid surface in such a manner that clear streaky patterns develop. Interpretation of the patterns is not always simple, particularly when flow reversals occur; the reason is that streak patterns do not indicate the flow direction. Photography and Lighting. Many kinds of cameras are used for flow visualization, including still, stereo, and cinematic (Adrian, 1986). The video camera has become increasingly popular due to advances in resolution, and it is possible to interface certain models with a computer to obtain quantitative data. Stroboscopes, as well as laser, tungsten, and xenon lighting, are commonly employed to visualize tracers. Depending on the application, images can be produced using steady light to generate streak lines; alternatively, images can be produced using light pulses which are triggered by the camera shutter or a stroboscope to freeze the motion. If two-dimensional visualization is desired, the flow field can be illuminated by creating a narrow sheet of light directed through a window in the test section. Fig. 13.23 on page 679 demonstrates the technique


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677

(a)

9 8 7

Y (CM)

6 5 4 3 2 1 0

0

2

4

6

8 X (CM)

10

12

14

16

(b) Fig. 13.21 Particle Image Velocimetry (PIV): (a) photograph of particle pathlines; (b) scaled velocity vectors. (Courtesy of R. Bouwmeester.)


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2 cm

(a)

(b)

(c)

Fig. 13.22 Molecular Tagging Velocimetry (MTV): (a) sample grid within a vortex ring just after laser tagging; (b) displaced grid 6 ms later; (c) velocity field of the vortex ring approaching a wall. (Courtesy of C. MacKinnon and M. Koochesfahani.)

of Laser Induced Fluorescence (LIF) to visualize a spanwise flow structure and mixing field in an unsteady wake flow field. Three-dimensional imaging can be accomplished by stereoscopic pictures or by use of holography.


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679

UO = 10 cm/s Re θ = 100

Laser sheet Spanwise flow structure and mixing field are examined using LIF

UO UO

y x 4 cm

z

8 cm CCD camera

Upper free-stream oscillates sinusoidally at F = 6 Hz and rms amplitude 10% of freestream speed

(a)

(b) Fig. 13.23 Laser Induced Fluorescence (LIF): (a) experimental layout; (b) detail of spanwise flow structure measured by LIF. (Courtesy of C. MacKinnon and M. Koochesfahani.)

13.4.2 Index of Refraction Methods The three index of refraction techniques are schlieren, shadowgraph, and interferometry. They are used in attempting to visualize flows when the density differences are either naturally or artificially altered; both gas and liquid mediums can be employed. The methods depend on the variation of the index of refraction in a transparent fluid medium and the resulting deflection of a light beam directed through the medium. Most often the flows to be studied are twodimensional, and the light beam is directed normal to the flow field; the result is a measurement integrated over the test section. The index of refraction is a function of the thermodynamic state of the fluid and typically depends only on the density. Examples of flows that are investigated with these techniques are high-speed flows with shock waves, convection flows, flame and combustion


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Chapter 13 / Measurements in Fluid Mechanics Light source Test section Converging mirror

Converging mirror

Knife edge Screen

(a)

(b)

Fig. 13.24 Schlieren system: (a) schematic; (b) helium jet entering atmospheric air. (Courtesy of R. Goldstein.)

flow fields, mixing of fluids of different densities, and forced-convection flows with heat or mass transfer. With a schlieren system, one measures the angle of the light beam after it has passed through the test section and emerges into the surrounding air (Fig. 13.24a); the angle is proportional to the first spatial derivative of the index of refraction. The light beam is turned in the direction of increasing index of refraction, or for most media, toward the region of higher density. A representative schlieren image is shown in Fig. 13.24b. A shadowgraph system measures the linear displacement of perturbed light (Fig. 13.25). In contrast to the schlieren imaging, the light pattern in a shadowgraph is determined by the second derivative of the index of refraction. Shadowgraph and schlieren systems are usually employed for qualitative flow visualization, whereas interferometers are often used in quantitative studies. Light source Test section

Screen Converging mirror

(a)

(b) Fig. 13.25 Shadowgraph system: (a) schematic; (b) helium jet entering atmospheric air. (Courtesy of R. Goldstein.)


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Sec. 13.5 / Data Acquisition and Analysis 681 Monochromatic light source

Lens

Mirror

Splitter plate

Test section

Splitter plate

Mirror

(b) (a)

Screen

(a)

Fig. 13.26 Mach–Zehnder interferometer: (a) schematic; (b) flow over heated gasturbine blades. (Courtesy of R. Goldstein.)

An interferometer responds to differences in optical path length; since a flow field with a variable density represents an optical disturbance, the phase of transmitted light waves is changed relative to undisturbed waves. A two-beam interferometer is designed to measure that phase difference (Fig. 13.26a). When the disturbed and undisturbed light beams are recombined, interference fringes appear on the screen that are a measure of the density variation in the flow field. An example of an interferogram is shown in Fig. 13.26b.

13.5 DATA ACQUISITION AND ANALYSIS This section begins with a brief introduction to automated acquisition of flow data using personal computers. Next, methodology to deal with the uncertainty of data is presented, including the manner in which those uncertainties are propagated when one flow parameter is represented indirectly as the result of measuring others. Finally, since a number of flow relations involve exponential formulations, a means of curve fitting exponential data is presented.

13.5.1 Digital Recording of Data Some of the measurements that have been described in previous sections are made simply by the direct reading of an instrument; those readings are probably steady-state, time-averaged, or perhaps root-mean-square values. Examples include reading a manometer attached to an orifice flowmeter, monitoring a digital rpm readout of a vane anemometer, and recording the pressure indicated on a Bourdon gage. The data acquired in this manner are usually sufficiently


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accurate for the desired goal; indeed, any further sophistication used to collect the data could be costly and unnecessary. On the other hand, there are situations that require more complex means of data acquisition. For instance, monitoring transient pressures in a pipeline using a strain-gage transducer, or observing turbulent velocity fluctuations in a free-stream flow, require that data be recorded continuously and accurately. Another illustration is the capture of the movement of a large number of moving particles in a two-dimensional plane in order to determine an unsteady flow field. Such measurements necessitate that data be recorded automatically, rapidly, and accurately. The advent of the personal computer has revolutionized automated acquisition of flow data. Many of the transducers in use provide a voltage output signal (an analog signal) that must be converted into a digital signal before the computer can store and process the data. Readily available today are a number of plug-in analog input/output (I/O) boards specifically made for personal computers, which can be inserted into the central processing unit bus. These I/O boards consist of input multiplexers, an input amplifier, a sample-and-hold circuit, one or more output digital-to-analog converters, and interfacing and timing circuitry. In addition, some include built-in random-access memory, read-only memory, and real-time clocks. Compatible software is available so that programming either is unnecessary or requires minimal effort.

13.5.2

Uncertainty Analysis

When flow measurements are made, it is important to realize that no recorded value of a given parameter is perfectly accurate. Instruments do not measure the so-called “true value” of the parameter, but they will provide an estimate of that value. Along with each measurement, one must attempt to answer the important question: How accurate is that measurement? The inaccuracies associated with measurements can be considered as uncertainties rather than errors. An error is the fixed, unchangeable difference between the true value and the recorded value, whereas an uncertainty is the statistical value that an error may assume for any given measurement. The purpose of this section is to describe a technique for propagating uncertainties of flow measurements into results. It has a variety of uses, among them helping us to decide whether calculations agree with collected data or lie beyond acceptable limits. Perhaps more importantly, it can help us to improve a given experiment or measuring procedure. It is not meant to be a substitute for accurate calibrations and sound experimental techniques; rather, it should be viewed as an additional aid for obtaining good, reliable results. Only a brief introduction is presented here; the reading of Coleman and Steele (1989) and Kline (1985) is recommended for a more thorough treatment of the subject. Uncertainty can be divided into three categories: (1) uncertainty due to calibration of instruments, (2) uncertainty due to acquisition of data, and (3) uncertainty as a result of data reduction. The total uncertainty is the sum of bias and precision. Bias is the systematic uncertainty that is present during a test; it is considered to remain constant during repeated measurements of a certain set of parameters. There is no statistical formulation that can be applied to estimate bias, hence its value must be based on estimates. Calibrations and independent


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Sec. 13.5 / Data Acquisition and Analysis 683

measurements will aid in its estimation. Precision is alternatively termed repeatability; it is found by taking repeated measurements from the parameter population and using the standard deviation as a precision index. For the situation in which repeated measurements are not taken (which occurs quite often for flow measurements), a single value must be used; however, less precision will be obtained. Fig. 13.27 illustrates how uncertainties are defined. A sampling of data (solid line) is shown that is assumed to be distributed in a Gaussian manner. The parameter xi is one particular measured variable, or measurand, <xi> is the average of all the measured variables, and dxi is the uncertainty interval, or precision, associated with xi. The uncertainty for the measurand can be given as (13.5.1)

xi < xi > dxi

Furthermore, it is necessary to state the bias and the probability P of recording the measurand with the stated precision. In addition to defining uncertainties in the data, it is necessary to propagate those uncertainties into the results. A two-step process is described as follows. 1.

Describe the uncertainty for the measurands. Each measurand is assumed to be an independent observation, and it must come from a Gaussian population. Also, the precision for each measurand must be quoted at the same probability P. For example, consider an experiment to calibrate a weir, in which the measurands are the water level upstream of the weir and the discharge. The water level is measured using a water-level gage, and the discharge is recorded by measuring the time it takes to fill a known volume. Thus the two measurands are independent. In addition, based on experience and perhaps some repeated measurements, one can reasonably state that (1) there is no bias and (2) the recorded precision for each measurand has a probability of P 0.95; that is, 20 out of every 21 readings of that measurand would yield the recorded value plus or minus the stated precision.

Average of all measured values

Frequency of measurement

True value

Precision

δx i

Bias < xi >

xi

Measured parameter

Fig. 13.27

Uncertainties associated with a sampling of data.


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2.

Compute the overall uncertainty interval. Let the parameter R represent the result computed from n variables x1, x2, . . . , xn; that is, R R(x1, x2, . . . , xn). Here we assume that R is continuous and has continuous derivatives in the desired domain. In addition we assume that the variables are uncorrelated and the uncertainty intervals are uncorrelated; that is, they are independent of one another. Then the overall uncertainty in the result dR can be obtained by expanding R in a Taylor series about the measurands x1, x2, . . . , xn and obtaining the rootsum-square result dR

R dx1

x1

冤冢

2

R dx2

x2

冣冢

2

冣

R dxn

xn

冢

2 1/2

冣冥

(13.5.2)

Situations arise when the expression for R becomes so complicated that it is difficult to obtain the partial derivatives in Eq. 13.5.2. In such a case, one can employ finite difference approximations of the partial derivatives to simplify the analysis. For example, consider a forward finite difference approximation for the first partial derivative: R(x1 dx1, x2, . . . , xn) R(x1, x2, . . . , xn)

R ⬇ (x1 dx1) x1

x1

(13.5.3)

Here we have assumed that in the denominator the finite difference interval is equivalent to the uncertainty interval dx1. The numerator of Eq. 13.5.3 is the difference between R evaluated at x1 dx1 and x1 with the remaining variables fixed. Expressing the other partial derivatives in Eq. 13.5.2 in a similar manner and substituting the results back into the relation we have the desired result:

dR ⬇

冦

冧

1/2

[R(x1 dx1, x2, . . . , xn) R(x1, x2, . . . , xn)]2 [R(x1, x2 dx2, . . . , xn) R(x1, x2, . . . , xn)]2 [R(x1, x2, . . . , xn dxn) R(x1, x2, . . . , xn)]

(13.5.4)

2

Equation 13.5.4 can easily be evaluated using a spreadsheet algorithm. Use of either Eq. 13.5.2 or 13.5.4 provides an estimate of dR. The overall uncertainty interval is an estimate of the standard deviation of the population of all possible experiments like the one being conducted.

Example 13.1 Discharge and pressure data are collected for water flow through an orifice meter in the pipe of Fig. E13.1a. The orifice diameter is 35.4 mm, and the diameter of the pipe is 50.8 mm. The original data are reduced to the form shown in the accompanying table. The second column is the discharge Q; the precision dQ (zero bias, P 0.95) is given in the third column; and the pressure head data h h1 h2 are presented in the


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fourth column. In addition, the precision for each of the pressure measurements is determined to be the same, namely, d(h) 0.025 m (zero bias, P 0.95). The fifth and sixth columns show the Reynolds numbers and the flow coefficient K computed using Eq. 13.3.8. Determine the uncertainties in the result K as a result of the estimated uncertainties in the measurands Q and h. Δh = h1 – h2 = (p1 – p2)/γ p2

p1

Q

Fig. E13.1a Solution Equation 13.3.8 is used in conjunction with Eq. 13.5.4 to propagate the uncertainties from Q and h into K. For example, consider the data for i 6. Calculation of dK proceeds as follows. The orifice area is p A0 0.03542 9.84 104 m2 4

Data no. i

Q (m3/s)

dQ (m3/s)

h (m)

Re

K

dK

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0.00119 0.00162 0.00200 0.00223 0.00251 0.00276 0.00294 0.00314 0.00330 0.00346 0.00373 0.00382 0.00408 0.00424 0.00480

0.00010 0.00012 0.00015 0.00015 0.00015 0.00017 0.00018 0.00019 0.00016 0.00017 0.00017 0.00017 0.00018 0.00019 0.00021

0.139 0.265 0.403 0.517 0.655 0.781 0.907 1.008 1.134 1.247 1.386 1.512 1.663 1.852 2.293

42 800 58 300 71 900 80 200 90 300 99 300 105 700 112 900 118 700 124 400 134 200 137 400 146 700 152 500 172 600

0.732 0.722 0.723 0.711 0.711 0.716 0.708 0.717 0.711 0.711 0.727 0.713 0.726 0.715 0.727

0.085 0.062 0.058 0.051 0.045 0.046 0.045 0.045 0.035 0.036 0.034 0.032 0.033 0.032 0.032

and the diameter ratio is 35.4 b 0.697 ⯝ 0.7 50.8 With Eq. 13.3.8, values of K are determined for (Q, h), (Q dQ, h), and (Q, h d(h)): (continued)


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0.00276 K(Q, h) 0.716 4 9.84 10 兹2苶81 9.苶.781 0苶 0.00276 0.00017 K(Q dQ, h) 9.84 104 兹2苶81 9.苶.781 0苶 0.761 0.00276 K(Q, h d(h)) 0.705 4 9.84 10 兹2苶81 9.苶0.781 (苶苶5) 0.02苶 These are then substituted into Eq. 13.5.4, with the variable K replacing the result R in that relation: dK {[K(Q dQ, h) K(Q, h)]2 [K(Q, h d(h)) K(Q, h)]2}1/2 [(0.761 0.716)2 (0.705 0.716)2]1/2 0.046 The results for all measurands are similarly evaluated and shown in the last column in the table. In addition, they are plotted in Fig. E13.1b, where the vertical lines, or uncertainty bands, associated with each value of K have the magnitude of 2dK. The bands can be interpreted as the estimated precision for K, based on zero bias and P 0.95 for the two measurands. Note that the precision dK stabilizes at approximately 0.032 at the higher Reynolds numbers and that the magnitude of dK is greater at lower Reynolds numbers. An alternative way to express this is that the uncertainty of K decreases with increasing Reynolds number. The data in the figure should be compared with the orifice curve labeled b 0.7 in Fig. 13.10. 1.0

0.9

0.8 K 0.7

0.6

0.5 4

6

8

10

20

Re × 10 −4

Fig. E13.1b


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A Mathcad solution of Example 13.1 is shown in Fig. 13.28.

Input data: D := 0.0354

Q :=

ORIGIN := 1

N := 15

v := 1.006 106

g := 9.81

0.00119

0.00010

0.139

0.025

0.00162

0.00012

0.265

0.025

0.002

0.00015

0.403

0.025

0.00223

0.00015

0.517

0.025

0.00251

0.00015

0.655

0.025

0.00276

0.00017

0.781

0.025

0.00294

0.00018

0.907

0.025

0.00314

dQ :=

0.00019

¢ h :=

1.008

d¢h :=

0.025

0.0033

0.00016

1.134

0.025

0.00346

0.00017

1.247

0.025

0.00373

0.00017

1.386

0.025

0.00382

0.00017

1.512

0.025

0.00408

0.00018

1.663

0.025

0.00424

0.00019

1.852

0.025

0.0048

0.00021

2.293

0.025

Define function to compute K:

q

K(q, ¢ H) := a

. D2 4

b

. (2.g .¢H) 0.5

Compute uncertainty in K: i := 1 ..N KKi := K(Qi, ¢ hi) dKi := 2(K(Qi dQi, ¢hi) K(Qi, ¢hi))2 (K(Qi, ¢hi d¢hi) K(Qi, ¢hi))2 Rei :=

4 . Qi .v.D

Fig. 13.28 Mathcad Solution to Example 13.1.


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Fig. 13.28 (continued)

13.5.3 Regression Analysis A significant number of flow-measuring devices correlate the discharge with pressure drop or head according to the general formula Y CX m

(13.5.5)

in which Y is the derived result, X is the measurand, and C and m are constants. By taking a number of independent measurements of X and Y, estimates of C and m can be found. A systematic approach to determine C and m is to find the best estimates based on the method of least squares. If the logarithm of each side of Eq. 13.5.5 is taken, it becomes ln Y ln C m ln X

(13.5.6)

which is of the form


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y b mx

(13.5.7)

By association we observe that Eq. 13.5.6 is linear provided that y ln Y, b ln C, and x ln X. Consider the data set Xi, Yi, i 1, . . . , n. The objective is to generate a straight line through the logarithms of the data (xi, yi) such that the errors of estimation are small. That error is defined as the difference between the observed value yi and the corresponding value on the line; in Fig. 13.29, si is the error associated with the data point (xi, yi). The method of least squares requires that to minimize the error, the sum of the squares of the errors of all the data be as small as possible. From Eq. 13.5.7, n

S 兺 s 2i i1 n

兺 [ yi (b mxi)]2 i1

(13.5.8)

The parameter S is minimized by forming the partial derivatives of S with respect to b and m, setting the resulting algebraic equations to zero, and solving for b and m. It is left as an exercise for the reader to derive the results: xiyi ( xi yi)앛n m x 2i ( xi)2앛n

(13.5.9)

yi m xi b n

(13.5.10)

C eb

(13.5.11)

A simple computer algorithm can be prepared to evaluate m and C based on a set of input data (Xi, Yi), or alternatively, a spreadsheet solution can be used.

y

si (x i , y i )

x

Fig. 13.29 Method of least squares.


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Example 13.2 With the data given in Example 13.1, perform a least squares regression to determine the coefficient C and exponent m in Eq. 13.5.5. Compare the result with Eq. 13.3.8. Solution: Equations 13.5.9 to 13.5.11 are employed to evaluate C, b, and m. In Eqs. 13.5.9 and 13.5.10, x is replaced by ln h ln(h1 h2), and y is replaced by ln Q. The calculations are tabulated in the accompanying table.

i

xi ln h

yi ln Q

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1.9733 1.3280 0.9088 0.6597 0.4231 0.2472 0.09761 0.007968 0.1258 0.2207 0.3264 0.4134 0.5086 0.6163 0.8299

6.7338 6.4253 6.2146 6.1058 5.9875 5.8925 5.8293 5.7635 5.7138 5.6665 5.5913 5.5675 5.5017 5.4632 5.3391

2.5886

87.7954

x 2i 3.8939 1.7636 0.8259 0.4352 0.1790 0.06111 0.009528 0.00006349 0.01583 0.04871 0.1065 0.1709 0.2587 0.3798 0.6887

xiyi 13.2878 8.5328 5.6478 4.0280 2.5333 1.4566 0.5690 0.0459 0.7188 1.2506 1.8250 2.3016 2.7982 3.3670 4.4309

8.8374

19.3173

Substitute the summed values into Eqs. 13.5.9 to 13.5.11: 19.3173 (2.5886)(87.7954)/15 m 0.497 8.8374 (2.5886)2/15 87.7954 0.497(2.5886) b 5.7673 15 C exp(5.7673) 0.00313 Hence the discharge is correlated to h by the regression curve Q 0.00313 h0.497 Based on the analysis performed in Example 13.1, the average value of K is 0.718. The area of the orifice is A0 9.84 104m2. Substituting these into Eq. 13.3.8, we find that Q 0.00313 h0.500 Thus the two results agree quite closely. The data and regression line are plotted in Fig. E13.2.


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0.005 0.004

0.003 Q (m3/s) 0.002 Q = 0.00313 Δh 0.497

0.001 0.1

0.2

0.4

0.6 0.8 1.0 Δh (m)

2.0

Fig. E13.2

A Mathcad solution for Example 13.2 is provided in Fig. 13.30.

Input data: n := 15

0.00119 0.139 0.00162 0.265 0.002

0.403

0.00223 0.517 0.00251 0.655 0.00276 0.781 0.00294 0.907 0.00314 1.008 0.0033

1.134

0.00346 1.247 0.00373 1.386 0.00382 1.512 0.00408 1.663 0.00424 1.852 0.0048

Fig. 13.30

2.293

Mathcad solution to Example 13.2.


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Chapter 13 / Measurements in Fluid Mechanics Convert to logarithmic form:

X := ln(M ) Y := ln(M )

Compute m, b, and C:

m :=

n1 n1 1 n1 a a Xi # Yi b # a Xi # a Yi n i0 i0 i0 n1

n1

2

2 #1 a (Xi) a a Xi b n i0 i0

b = 5.767

m = 0.496

n1

n1

# a Yi m a Xi

b :=

i0

i0

n

C := exp(b)

C = 3.128 103

Plot Q versus ¢ h:

Discharge (cubic meters per second)

1.01

M á0á C .(M á1á )

m

1.10−3 0.1

10

1 M

á1á

á1á

,M Head (meters)

Fig. 13.30 (continued)


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Problems 693

13.6 SUMMARY There are many devices available to measure flow parameters; we have presented some of the more common ones. The monitoring of pressure and velocity at discrete locations in flow fields was first discussed, followed by descriptions of integrated flow rate or discharge measurements. Various devices are used for measurement of these parameters, ranging from purely mechanical to electromechanical. Flow visualization is an important technique that is employed in both industrial and research laboratories to study complex flow fields, especially turbulent ones. Uses of tracers and index of refraction methods are presented. Of these methods, velocimetry techniques are rapidly gaining favor due to computer imaging and the development of laser-activated dyes. Today, a significant number of measurements are automated and may require analog or digital interfacing with data loggers or personal computers. Regardless of the manner in which data is acquired and stored, the data taker must be aware of uncertainties that are present and unavoidable in the process of acquiring data. We have shown how the uncertainties present in measured parameters can be propagated into results using spreadsheet or computational software analyses. These techniques are useful, for example, to determine how accurate and reliable a flow meter may be, or to design an experiment to calibrate a velocity meter that will insure a desired degree of accuracy. Finally, regression analysis is applied to determine the parameters that describe an exponential relationship that correlates discharge with pressure drop or head.

PROBLEMS 13.1

13.2

13.3

13.4

A 20° inclined manometer attached to a piezometer opening is used to measure the pressure at the wall of airflow. If the reading is measured to be 4 cm of mercury, calculate the pressure at the wall. A pitot-static probe is to be used for repeated measurements of the velocity of an atmospheric airstream. It is desired to relate the velocity V to the manometer reading h in centimeters of mercury (i.e., V C 兹h 苶 ). Calculate the value of the constant C for a laboratory at 20°C situated at: (a) Sea level (b) 2000 m elevation A pitot-static probe attached to a manometer that has water as the manometer fluid is proposed as the measuring device for an airflow with a velocity of 8 m/s. What manometer reading is expected? Comment as to the use of the proposed device. Water at 75°F flows through a 0.1-in.-diameter tube into a calibrated tank. If 2 quarts are collected in 10 minutes, calculate the flow rate (ft3/sec), the mass

13.5

flux (slug/sec), and the average velocity (ft/sec). Is the flow laminar, turbulent, or can’t you tell? A velocity traverse in a rectangular channel that measures 10 cm by 100 cm is represented by the following data. Estimate the flow rate and the average velocity assuming a symmetric plane flow.

y (cm)

0

0.5

1

2

3

4

5

u (m/s)

0

5.2

8.1

9.2

9.8

10

10

13.6

A velocity traverse in a 10-cm-diameter pipe is represented by the following data. Estimate the flow rate and the average velocity assuming a symmetric flow.

r (cm)

0

1

2

3

4

4.5

u (m/s)

10

10

9.8

9.2

8.1

5.2


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694


13.7

The flow rate of water in a 12-cm-diameter pipe is measured with a 6-cm-diameter venturi meter to be 0.09 m3/s. What is the expected deflection on a water-mercury manometer? Assume a water temperature of 20°C. The flow rate of 20°C water flow in a 24-cmdiameter pipe is desired. If a water-mercury manometer reads 12 cm, calculate the discharge if the manometer is connected to: (a) A 15-cm-diameter orifice (b) A 15-cm-diameter nozzle Calculate the flow rate of 40°C water in the pipes shown in Fig. P13.9.

13.8

13.9

6-cm-dia. throat

12 cm dia.

12 cm Hg (a) 6-cm-dia. throat

12 cm dia.

√1

2 (p 莦 p ) 冪莦 r T

1

in which √1 is the local velocity at location 1, r the density of the liquid, pT the total pressure measured by the tube, and p1 the static pressure at location 1. The density is estimated to be 680 50 kg/m3 (P 0.95, 0 bias). (a) Assume that the two pressures are measured separately using individual pressure gages: pT 102 1 kPa (P 0.95, 0 bias), p1 95 1 kPa (P 0.95, 0 bias). Estimate the uncertainty for √1. (b) Assume that the pressure difference pT p1 is measured using a single differential pressure gage: pT p1 7 1 kPa (P 0.95, 0 bias). Estimate the uncertainty for √1. (c) Which arrangement, (a) or (b), is preferable? 13.12 Experimental data used to calibrate discharge of water in a venturi meter are tabulated below. The diameter of the meter throat is 33.3 mm and the diameter at the upstream tap location of the differential manometer is 54.0 mm. The precision dQ of the discharge measurements has zero bias and P 0.95, and the precision for all the manometer measurements is d(h) 2 mm Hg, with zero bias and P 0.95. Note that the manometer readings are given in mm Hg. With use of Eq. 13.3.8, determine the uncertainties in K as a result of the uncertainties in the two measurands. Plot the results in a fashion similar to those in Example 13.1, and compare the curve with Fig. P13.10.

12 cm Hg (b)

Fig. P13.9 13.10

A pressure difference across an elbow in a water pipe is measured. Estimate the flow rate in the pipe. The pressure difference, pipe diameter, and radius of curvature are, respectively: (a) 80 kPa, 10 cm, 20 cm (b) 80 kPa, 5 cm, 20 cm (c) 10 psi, 4 in., 8 in (d) 10 psi, 2 in., 8 in Use a water temperature of 20°C. 13.11 A pitot tube is used to measure the velocity of gasoline flowing in a pipe. The estimate of the velocity is given by the equation

Data no. i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Q (m3/s)

dQ (m3/s)

h (mm Hg)

Re

0.00168 0.00208 0.00249 0.00281 0.00328 0.00355 0.00372 0.00402 0.00415 0.00444 0.00479 0.00493 0.00523 0.00533 0.00509

0.00011 0.00014 0.00016 0.00019 0.00022 0.00024 0.00025 0.00023 0.00024 0.00025 0.00027 0.00028 0.00030 0.00035 0.00029

13 22 32 40 54 62 70 82 91 100 113 123 134 140 140

64 200 79 500 95 100 107 400 125 300 135 600 142 100 153 600 158 600 169 700 183 000 188 400 199 800 203 700 194 500


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Problems 695 13.13

With the numerical results obtained from Example 13.1, plot the relative uncertainty dK/K versus Re on a linear scale. What conclusions can you draw? 13.14 Using the data from Problem 13.12, perform a least squares regression and determine the coefficient C and exponent m in Eq. 13.5.5. Compare the result with Eq. 13.3.8. Note that the manometer readings are given in mm Hg. 13.15 Establish a depth–discharge relation of the form Q CYm for a 60° V-notch sharp-crested weir using the data in the following table. In the equation, Y is the vertical distance from the weir notch to the undisturbed upstream water surface and Q is the discharge (see Fig. 10.11). Compare the result with Eq. 10.4.27.

Data no. i 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Q (ft3앛sec)

Y (ft)

0.1390 0.1300 0.1280 0.1140 0.0963 0.0937 0.0809 0.0781 0.0751 0.0697 0.0518 0.0435 0.0418 0.0319

0.387 0.378 0.374 0.357 0.336 0.320 0.312 0.306 0.302 0.297 0.263 0.244 0.238 0.206


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A CFD analysis of the cylinder of a proposed engine design. Left is the grid with areas of refined grid in critical flow areas. Right is a plot of the velocity vectors showing the flow field in a plane cross section. (Courtesy of Bassem Ramadan.)


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14 Computational Fluid Dynamics Outline 14.1 Introduction 14.2 Examples of Finite-Difference Methods 14.2.1 Discretization of the Domain 14.2.2 Discretization of the Governing Equations 14.2.3 Define Solution Algorithm 14.2.4 Comments on Choice of Difference Operators 14.3 Stability, Convergence, and Error 14.3.1 Consistency 14.3.2 Numerical Stability 14.3.3 Convergence 14.3.4 Numerical Errors 14.4 Solution of Couette Flow 14.5 Solution of Two-Dimensional, Steady-State Potential Flow 14.6 Summary

Chapter Objectives The objectives of this chapter are to: ▲ ▲ ▲ ▲ ▲

Present an introduction to finite-difference methods. Present ideas on consistency, numerical stability, convergence, and errors. Present methods of finite differences. Present a finite-difference solution to transient couette flow. Present a finite-difference solution to steady-state potential flow.

697


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Chapter 14 / Computational Fluid Dynamics

14.1 INTRODUCTION CFD solutions pages 669–672, 807, 821

The equations governing fluid-flow problems are the continuity, the Navier– Stokes, and the energy equations. These equations, derived in Chapter 5, form a system of coupled quasi-linear partial differential equations (PDEs). Because of the nonlinear terms in these PDEs, analytical methods can yield very few solutions. In general, analytical solutions are possible only if these PDEs can be made linear, either because nonlinear terms naturally drop out (e.g., fully developed flows in ducts and flows that are irrotational everywhere) or because nonlinear terms are small when compared to other terms so that they can be neglected (e.g., flows where the Reynolds number is less than unity). Yih (1969) and Schlichting & Gersten (2000) describe most of the better-known analytical solutions. If the nonlinearities in the governing PDEs cannot be neglected, which is the situation for most engineering flows, then numerical methods are needed to obtain solutions. Computational fluid dynamics, or simply CFD, is concerned with obtaining numerical solutions to fluid-flow problems by using the computer. The advent of high-speed and large-memory computers has enabled CFD to obtain solutions to many flow problems, including those that are compressible or incompressible, laminar or turbulent, chemically reacting or nonreacting, single- or multi-phase. Of the numerical methods developed to address equations governing fluid-flow problems, finite-difference methods (FDMs) and finite-volume methods (FVMs) are the most widely used. In this chapter, we give an introduction to the FDMS and present the solutions to two classical fluid-flow. Problems using finite differences (FD). Since computers are used to obtain solutions, it is important to understand the constraints that they impose. Of these constraints, four are critical. The first of these is that computers can perform only arithmetic (i.e., , , , and ) and logic (i.e., true and false) operations. This means non-arithmetic operations, such as derivatives and integrals, must be represented in terms of arithmetic and logic operations. The second constraint is that computers represent numbers using a finite number of digits. This means that there are round-off errors and that these errors must be controlled. The third constraint is that computers have limited storage memories. This means solutions can be obtained only at a finite number of points in space and time. Finally, computers perform a finite number of operations per unit time. This means solution procedures should minimize the computer time needed to achieve a computational task by fully utilizing all available processors on a computer and minimizing the number of operations. With these constraints, FD methods generate solutions to PDEs through the following three major steps: 1.

Discretize the domain. The continuous spatial and temporal domain of the problem must be replaced by a discrete one made up of grid points or cells and time levels. The ideal discretization uses the fewest number of grid points/cells and time levels to obtain solutions of desired accuracy.


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Sec. 14.2 / Examples of Finite-Difference Methods 699

2.

3.

Discretize the PDEs. The PDEs governing the problem must be replaced by a set of algebraic equations with the grid points/cells and the time levels as their domain. Ideally, the algebraic equations—referred to as finite-difference equations (FDEs) should describe the same physics as those described by the governing PDEs. Specify the algorithm. The step-by-step procedure by which solutions at each grid points/cells are obtained from the FD equations when advancing from one time level to the next must be described in detail. Ideally, the algorithm should ensure not only accurate solutions but also efficiency in utilizing the computer.

These steps are illustrated through example problems in Sections 14.4 and 14.5.

14.2 EXAMPLES OF FINITE-DIFFERENCE METHODS To illustrate FDMs, consider the unsteady, incompressible, laminar flow of a fluid with constant kinematic viscosity n between two parallel plates separated by a distance H, as shown in Fig. 14.1. Initially, both plates are stationary, and the fluid between them is stagnant. Suddenly at time t 0, the lower plate moves horizontally to the right (in the positive x-direction) at a constant speed of V0. The equations governing this flow are the continuity equation and the x- and y-component Navier–Stokes equations. Since the flow is parallel (i.e., √ 0) and the pressure is the same everywhere, these equations reduce to the following single linear PDE: u 2u n 2 t y

(14.2.1)

The initial and boundary conditions for Eq. 14.2.1 are, assuming u u(y, t), u(y, t 0) 0,

u(0, t) V0

u(y H, t) 0 (14.2.2)

u( y, 0) 0

u(H, t) 0

u(0, t) V0

y

ρ, ν

H x

V0

Fig. 14.1 Flow between stationary and moving parallel plates.


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Since Eq. 14.2.1 is linear and the boundary conditions given by Eq. 14.2.2 are homogeneous, separation of variables is able to provide the exact analytical solution 1 u/V0 1 y/H 2 兺 sin (npy/H) exp (n2p 2nt/H 2) n1 np

(14.2.3)

This exact solution can be used to evaluate the accuracy of FDMs to be presented. As mentioned, three steps are involved in generating solutions to PDEs by using an FDM. These three steps are illustrated below, one at a time for Eqs. 14.2.1 and 14.2.2.

14.2.1 Discretization of the Domain To discretize the domain, we note that the spatial domain is a line segment between 0 and H, and that the temporal domain is a ray emanating from t 0. Though the temporal domain is of infinite extent, the duration of interest is finite, say from t 0 to t T, where T is the time when steady state is achieved. Here, the spatial domain, 0 y H, is discretized by replacing it with J equally distributed stationary grid points, and the temporal domain, 0 t T is discretized by replacing it with equally incremented time levels (see Fig. 14.2). This discretization of the domain is one of many instances. For example, the grid points do not have to be uniformly distributed nor do they have to be stationary. Each point in the discretized domain, shown in Fig. 14.2, has coordinates ( yj, t n), given by yj ( j 1) y, t n n t,

j 1, 2, 3, . . . , J

(14.2.4)

n 0, 1, 2, . . .

(14.2.5)

t

n+1

H yj = (j−1) Δy, Δy = —— J−1

Δt

n •••

tn = n Δ t

2 1 0

Δy

y 1 2 y=0

Fig. 14.2

•••

j−1 j j + 1 • • •

J−1 J y=H

Grid system and time levels for problem depicted in Figure 14.1.


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where y H/(J 1) is the distance between two adjacent grid points (or the grid spacing), and t is the time-step size. The sizes of the grid spacing and the time step depend on the length and time scales that need to be resolved and on the properties of the discretized PDEs used to obtain solutions. The solution sought—u( y, t) in Eq. 14.2.1—will be obtained only at the grid points and the time levels. That solution is denoted as u nj u(yj, t n)

(14.2.6)

where subscripts denote the locations of the grid points and superscripts denote time levels.

14.2.2 Discretization of the Governing Equations With the domain discretized, the next step is to replace the PDEs governing the problem by a set of algebraic or FDEs that use the grid points and time levels as the domain. With FDMs, PDEs are discretized by replacing derivatives with difference operators. Thus, discretization involves two parts. First, derive difference operators. Then, select difference operators. Derive Difference Operators. There are many different ways to derive difference operators. For problems with smooth solutions (i.e., solutions without discontinuities such as shock waves), a method based on the following theorem is often used: If a function, u, and its derivatives are continuous, single-valued, and finite, then the value of that function at any point can be expressed in terms of u and its derivatives at any other point by using a Taylor series expansion, provided the other point is within the radius of convergence of the series. With this theorem, uj1 and uj1 can be expressed in terms of uj and its derivatives as

冢冣冢冣冢冣 u u y u y u 冢冣 y 冢冣冢冣 y y 2! y 3!

u 2u y2 3u y3 uj1 uj y 2 3 y j y j 2! y j 3! 2

uj1

j

2

3

2 j

j

(14.2.7)

3

3 j

(14.2.8)

From the above two Taylor series, we can readily derive the following four difference operators by solving for (u/y)j and (2u/y2)j either directly or by adding or subtracting the two equations: uj1 uj

O(y) 冢uy 冣 y

(14.2.9)

j

uj uj1

冢uy 冣 y O(y)

(14.2.10)

j

uj1 uj1

冢uy 冣 2y O(y )

(14.2.11)

uj1 2uj uj1 2u 2 O(y2) j y y2

(14.2.12)

2

j

冢冣


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In the above difference operators, O(y) and O(y2) denote the truncation errors (i.e., terms in the Taylor series that have been truncated). In Eqs. 14.2.9 and 14.2.10, the power of y in O(y) is one because the leading term of the truncation errors for these two difference operators is multiplied by y raised to the first power. Such difference operators are said to be first-order accurate. In Eqs. 14.2.11 and 14.2.12, that power is two because the leading term in the truncation errors is multiplied by y raised to the second power. Such difference operators are said to be second-order accurate. The higher the order of accuracy, the greater is the number of terms retained in the truncated Taylor series. For sufficiently small y, a higher-order accurate difference operator is a more accurate representation of a smooth function than a lower-order accurate difference operator. In general, second-order accuracy is adequate. The grid points used to construct a difference operator form the stencil of that difference operator. If the stencil for a difference operator at yj involves only grid points with indices greater than or equal to j such as Eq. 14.2.9, then that difference operator is said to be a forward-difference operator. If grid points all have indices less than or equal to j such as Eq. 14.2.10, then that difference operator is said to be a backward-difference operator. If the number of grid points behind and ahead of j are exactly the same such as Eqs. 14.2.11 and 14.2.12, that difference operator is said to be a central-difference operator. If the number of grid points ahead of and after j are not the same, that difference operator is said to be either biased backward or biased forward. Taylor series can be used to derive difference operators for any derivative to any order of accuracy and using any stencil. To illustrate, consider the derivation of a difference operator for a first-order derivative, (u/y)j, with the following specifications: third-order accurate, O(y3), with a stencil that includes only one downstream grid point. The derivation of this difference operator involves the following four steps: Step 1: Determine the number of terms to be kept in each Taylor series. That number, denoted as N, is equal to the order of the derivative for which a difference operator is sought plus the order of accuracy desired. For this case, the order of the derivative is 1, and the order of accuracy desired is 3. Thus, N is 4. Step 2: Decide on a stencil based on N points for the difference operator at point j. For this case with N 4 and u 0, the 4 points of the stencil are j 2, j 1, j, j 1. Other possible stencils include: ( j 3, j 2, j 1, j) or completely backward, ( j 1, j, j 1, j 2) or biased forward, and ( j, j 1, j 2, j 3) or completely forward. Step 3: Construct N 1 truncated Taylor series about uj from u at all points of the stencil except point j. Performing this operation gives

冢冣冢冣冢冣 u u y u y u 冢冣 y 冢冣冢冣 O(y ) y y 2! y 3! u u (2y) u (2y) u 冢冣 2y 冢冣冢冣 O(y ) y y 2! y 3!

u 2u y2 3u y3 uj1 uj y 2 3 O(y4) y j y j 2! y j 3! 2

uj1

j

2

j

3

4

j

2 j

2

uj2

3

3 j

2

3

(14.2.13)

3

4

j

2 j

(14.2.12)

3 j

(14.2.14)


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In the above three equations, the unknowns are the first, second, and third derivatives, and they must be expressed in terms of u at points in the stencil. Step 4: Solve N 1 coupled linear equations for the difference operator sought. For the first derivative, solving the above three equations gives 2uj1 3uj 6uj1 uj2

O(y ) 冢uy 冣 6y 3

(14.2.15)

j

Difference operators for the second and third derivatives are also derived in the process of solving for the above difference operator. The orders of accuracy of these difference operators are second order for the second derivative and first order for the third derivative. In Table 14.1, some commonly used difference operators are summarized. Note that if j and y are replaced by n and t, then the difference operators can also be used for time derivatives. TABLE 14.1

Summary of Commonly Used Difference Operators

Description

Finite-Difference

Stencil j-1

First Derivative: (u앛y)j central, O(y2)

(uj1 uj1)/2y

backward, O(y)

(uj uj1)/y

forward, O(y)

(uj1 uj)/y

backward, O(y2)

(3uj 4uj1 uj2)/2y

forward, O(y2)

(3uj 4uj1 uj2)/2y

j

j+1

Second Derivative: (2f앛y2) (u앛y)j, u f 앛y central, O(y2)

( fj1 2fj fj1)/y2

backward, O(y)

( fj 2fj1 fj2)/y2

forward, O(y)

( fj 2fj1 fj2)/y2


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Page 704


Select Difference Operators. To select difference operators, we apply the governing PDE to be discretized (Eq. 14.2.1) at an arbitrary interior grid point (a grid point not on the boundary) and at a time between t n and t n1:

冢

冣

u 2u n 2 t y

n

(14.2.16)

j

where n is n, n 1, or some value between n and n 1. In this equation, it is assumed that the solution at time t n is known and that the solution at t n1 is sought. This assumption is generally applicable because the solution is always known at the previous time level. For example, at the zeroth time level (n 0), the solution is the initial condition. An Explicit Method. If n n in Eq. 14.2.16, then all spatial derivatives are evaluated at time t n, the previous time level where the solution is known. If we choose the forward-difference operator given by Eq. 14.2.9 for (u/t) nj (except replace y and j with t and n) and the central-difference operator given by Eq. 14.2.12 for (2u/y2) nj , then Eq. 14.2.16 becomes u n1 u nj u nj1 2unj u nj1 j n O(t, y2) t y2

(14.2.17)

The above FDE is first-order accurate in time and second-order accurate in space as indicated by O(t, y2). In this equation, u n1 is the unknown sought, and j solving for it gives u n1 bu nj1 (1 2b)u nj bu nj1, j

b n t/y2

(14.2.18)

The FDE in the form of Eq. 14.2.17 or 14.2.18 can be applied at any interior grid point ( j 2, 3, . . . , J 1). The FDEs at the boundary grid points ( j 1 and J) are obtained by using boundary conditions (BCs). For this simple problem, the BCs, given by Eq. 14.2.2, readily yield the following FDEs: un1 u n1 V0 1

(14.2.19)

u nJ

(14.2.20)

u n1 J

0

Note that the FDEs given by Eqs. 14.2.17 and 14.2.20 have only one unknown in them, namely, u n1 , a consequence of setting n n in Eq. 14.2.16. j A method made exclusively of such FDEs is said to be explicit. This particular explicit method with a first-order accurate forward-time differencing is known as the Euler explicit scheme. An Implicit Method. If n n 1 in Eq. 14.2.16, then all spatial derivatives are evaluated at t n1, the new time level where the solution is unknown. If we choose the backward-difference operator given by Eq. 14.2.10 for (u/t) n1 j to couple t n1 to t n and the central-difference operator given by Eq. 14.2.11 for (2u/y2) n1 , then Eq. 14.2.16 becomes j


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Sec. 14.2 / Examples of Finite-Difference Methods 705 n1 u n1 u n1 u n1 u nj j1 2u j j1 j n O(t, y2) 2 y t

(14.2.21)

which can be rewritten as n1 n bu n1 bu n1 j1 (1 2b)u j j1 u j ,

b n t/y2

(14.2.22)

Similar to Eq. 14.2.18, the above equation is first-order accurate in time and second-order accurate in space. But it differs in that there are three unknowns, n1 u n1 , and u n1 j1 , u j j1 , instead of one in the FDE. This difference is a consequence of evaluating spatial derivatives at t n1 instead of t n. By applying Eq. 14.2.22 at every interior grid point and by using Eqs. 14.2.19 and 14.2.20 for the boundary grid points, we obtain the following system of equations: Ax b

(14.2.23)

where

A

1 2b b b 1 2b b

u n1 u n2 bV0 2 n1 b u3 u n3 n1 1 2b b u4 u n4 . . . , x . . , b . . . . . . . . . . u nJ2 b 1 2b b u n1 J2 b 1 2b u n1 u nJ1 J1

(14.2.24) In the above system of equations, the FDEs at j 1 and 2 and at j J 1 and J have been combined. A method in which each FDE contains more than one unknown—so that solutions of simultaneous equations are needed—is said to be implicit. This particular implicit method with a first-order accurate backwardtime differencing is known as the Euler implicit scheme. A Generalized Method. If n is a value between n and n 1, then Eq. 14.2.16 can be written as u n1 u nj u j u t t

冢冣

n1 j

u (1 u) t

冢冣 , 0 ex11_4 Optimization terminated successfully: Relative function value changing by less than OPTIONS.TolFun x = 15.1818 0.0098 0.0170 0.0072

FIGURE F.3 Example 11.4 MATLAB® solution: (a) main algorithm, (b) function

subroutine, (c) output. Note that the Optimization Toolbox is required.


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Appendix 761

Given data: L1: ⫽ 50 L 2 : ⫽ 100 L 3 : ⫽ 300 g : ⫽ 9.81

D1 : ⫽ 0.15 D2 : ⫽ 0.10 D3 : ⫽ 0.10 P : ⫽ 20 # 103

f1 : ⫽ 0.02 f2 : ⫽ 0.015 f3 : ⫽ 0.025

K1 : ⫽ 2 K2 : ⫽ 1 K3 : ⫽ 1

H 1 : ⫽ 10 H 2 : ⫽ 30 H 3 : ⫽ 15

Equivalent lengths: Leq2 : ⫽

D2 # K 2 f2

Leq1 : ⫽

D1 # K 1 f1

Leq3 : ⫽

D3 # K 3 f3

Resistance coefficients: R1 : ⫽

8 # f1 # (L 1 ⫹ Leq1) 8 # f2 # (L 2 ⫹ Leq2) 8 # f3 # (L 3 ⫹ Leq3) R : ⫽ R : ⫽ 2 3 g # p2 # D15 g # p 2 # D52 g # p2 # D53

Initial estimates of unknowns: Q1 : ⫽ 0.05

Q2 : ⫽ 0.05

Q3 : ⫽ 0.05

H B : ⫽ 20

Solve for unknowns: Given P ⫺ H B ⫽ R 1 # Q1 # 0Q1| 9800 # Q1 H B ⫺ H 2 ⫽ R 2 # Q2 # |Q2| H1 ⫹

H B ⫺ H 3 ⫽ R 3 # Q3 # |Q3| Q1 ⫺ Q2 ⫺ Q3 ⫽ 0

43.845 0.054 Find (HB, Q1, Q2, Q3) ⫽ ± ≤ 0.032 0.021 FIGURE F.4

Example 11.5 Mathcad solution



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Appendix

(a) clear; %Given Data global H R P; L=[50 100 300]; D=[0.15 0.10 0.10]; f=[0.020 0.015 0.025]; K=[2 1 1]; H=[10 30 15]; g=9.81; p=20000; %Evaluate equivalent lengths and resistence coefficients Le=D.*k./f; R=8*f.*(L+Le)./(g*pi^2*D.^5); %Initial estimates of unknowns x0=[HB Q1 Q2 Q3] x0=[20 0.05 0.05 0.05]; %Call function file g.m and solve for unknows options=optimset ('precondbandwidth', Inf); [x, fval] = fsolve('g', x0, options); x

(b) %Define each function as g(x)=0 function G = g(x); global H R P; G = [H(1) x(1) x(1) x(2)

+ – – –

p/(9800*x(2)) – x(1) – R(1)*x(2)*abs(x(2)); H(2) – R(2)*x(3)*abs(x(3)); H(3) – R(3)*x(4)*abs(x(4)); x(3) – x(4)];

(c) >> ex11_5 Optimization terminated successfully: Relative function value changing by less than OPTIONS.TolFun x = 43.8449 0.0538 0.0324 0.0214

FIGURE F.5 Example 11.5 MATLAB® solution: (a) main algorithm, (b) function

subroutine, (c) output. Note that the Optimization Toolbox is required.


A B 1 2 3 4 ΔH 5 Pipe 4 6 7 Loop 1 Pipe 3 Pipe 2 8 Pipe 1 9 10 11 12 13 14 15 16 Pipe 2 17 Loop 2 Pipe 8 Pipe 7 18 Pipe 6 19 20 21 22 23 24 25 26 Pipe 3 27 Loop 2 Pipe 5 28 Pipe 8 29 30 31 32 101.710 3.57E–04

–0.036 ΔQ =

25.440 32.898 44.251

4.87E–03

24.000 32.000 42.000 98.000 1.12E–03

ΔQ =

0.720 0.640 –1.470 –0.110 ΔQ =

200 400 300

FIGURE F.6

24.817 34.039 42.853 0.770 0.724 –1.530

ΔQ = –9.03E–04

318.000

–1.550

0.060 0.040 –0.070

ΔQ =

9.433 1.632 –0.494 –10.281

130.000 42.000 32.000 114.000

0.130 8.450 0.070 1.470 –0.040 –0.640 –0.190 –10.830

ΔQ =

–0.146

0.809 0.676 –1.632

0.290

1.43E–03

102.589

320.779

0.062 0.043 –0.071

–0.464

8.907 1.530 –0.519 –10.382

1.47E–03

316.759

0.063 0.043 –0.073

0.135 0.073 –0.035 –0.185

317.183

135.276 43.519 27.650 110.738

9.03E–04

228.309

3.968 25.098 135.276 63.968

ΔQ =

–0.054

0.787 0.736 –1.578

5.29E–04

102.941

25.098 34.325 43.519

ΔQ = –9.83E–05

0.031

9.150 1.578 –0.478 –10.219

133.466 42.853 28.823 111.617 0.133 0.071 –0.036 –0.186

137.352 44.251 28.101 111.075

Example 11.6 using Excel® (a) Solution

0.064 0.041 –0.074

0.137 0.074 –0.035 –0.185

ΔQ =

ΔQ = –3.44E–04

2.98E–03

20.000 –0.039 –0.787 –9.150 –10.230

2R|Q|

O

0.062 0.043 –0.072

0.134 0.072 –0.035 –0.185

–0.019 –0.062 –0.134 –0.319

Q

P

10:05 PM

500 300 400 300

ΔQ =

ΔQ = –2.48E–03

–0.020 –0.063 –0.135 –0.320

RQ|Q|

–0.206

3.899 24.817 133.466 63.899

Q

N Iteration 4

226.081

0.078

20.000 –0.038 –0.770 –8.907 –10.208

2R|Q|

M

11/30/10

231.783

–0.019 –0.062 –0.133 –0.319

RQ|Q|

–0.691

4.495 25.440 137.352 64.495

Q

222.000

20.000 –0.051 –0.809 –9.433 –10.399

2R|Q|

0.550

–0.022 –0.064 –0.137 –0.322

RQ|Q|

4.000 24.000 130.000 64.000

L

20.000 –0.020 –0.040 –0.060 –0.720 –0.130 –8.450 –0.320 –10.240

K Iteration 3

100 200 500 100

Q

2R|Q|

J

RQ|Q|

I

Q

H Iteration 2

R

G

F

E Iteration 1

D

C

67730_15_appendix_p732-772.qxd Page 763

Appendix 763


A 1 2 3 4 5 6 7 Loop 1 8 9 10 11 12 13 14 15 16 17 Loop 2 18 19 20 21 22 23 24 25 26 27 Loop 2 28 29 30 31 32 200 400 300

Pipe 3 Pipe 5 Pipe 8

0.06 0.04 –0.07

=SUM(F15:F19) =SUM(E16:E19)

=SUM(F26:F28)

=SUM(E26:E28)

ΔQ = =–E30/F30

=2*C26*ABS(D26) =2*C27*ABS(D27) =2*C28*ABS(D28)

=C26*D26*ABS(D26) =C27*D27*ABS(D27) =C28*D28*ABS(D28)

ΔQ = =–E21/F21

=2*C16*ABS(D16) =2*C17*ABS(D17) =2*C18*ABS(D18) =2*C19*ABS(D19)

=C16*D16*ABS(D16) =C17*D17*ABS(D17) =C18*D18*ABS(D18) =C19*D19*ABS(D19)

=D26+F32–F13 =D27+F32 =D28+F32–F23

=D16+F23–F13 =D17+F23–F32 =D18+F23 =D19+F23

=D6+F13 =D7+F13–F32 =D8+F13–F23 =D9+F13

Q

G

10:05 PM

0.13 0.07 –0.04 –0.19

=SUM(F6:F9)

=SUM(E5:E9)

ΔQ = =–E11/F11

=2*C6*ABS(D6) =2*C7*ABS(D7) =2*C8*ABS(D8) =2*C9*ABS(D9)

2R|Q|

RQ|Q| 20 =C6*D6*ABS(D6) =C7*D7*ABS(D7) =C8*D8*ABS(D8) =C9*D9*ABS(D9)

F

E Iteration 1

11/30/10

FIGURE F.6 (cont’d) Example 11.6 using Excel® (b) Spreadsheet formulas

500 300 400 300

Pipe 2 Pipe 8 Pipe 7 Pipe 6

–0.02 –0.06 –0.13 –0.32

Q

R

100 200 500 100

D

C

764

ΔH Pipe 4 Pipe 3 Pipe 2 Pipe 1

B

67730_15_appendix_p732-772.qxd Page 764

Appendix



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Page 765

Appendix 765

Given data: R 1 : ⫽ 100 R 3 : ⫽ 400 R 5 : ⫽ 400 R 7 : ⫽ 400 R 2 : ⫽ 500 R 4 : ⫽ 100 R 6: ⫽ 300 R 8: ⫽ 300

H A : ⫽ 50 QF : ⫽ 0.15 H B : ⫽ 30 QG : ⫽ 0.15

Initial estimates of unknowns: Q1 : ⫽ 0.50 Q2 : ⫽ 0.50 Q3 : ⫽ 0.50

Q4 : ⫽ 0.50 Q5 : ⫽ 0.50 Q6 : ⫽ 0.50

Q7 : ⫽ 0.50 Q8 : ⫽ 0.50

H C : ⫽ 35 H D : ⫽ 35 H E : ⫽ 35

H F : ⫽ 35 H G : ⫽ 35

Solve for unknowns: Given H A ⫺ H C ⫽ R 1 # Q1 # |Q1| H C ⫺ H D ⫽ R 2 # Q2 # |Q2| H D ⫺ H E ⫽ R 3 # Q3 # |Q3| H E ⫺ H B ⫽ R 4 # Q4 # |Q4|

Q1 ⫽ Q2 ⫹ Q6 Q2 ⫽ Q3 ⫹ Q8 H C ⫺ H F ⫽ R 6 # Q6 # |Q6| Q3 ⫽ Q4 ⫹ Q5 H F ⫺ H G ⫽ R 7 # Q7 # |Q7| Q6 ⫽ Q7 ⫹ QF H D ⫺ H G ⫽ R 8 # Q8 # |Q8| Q7 ⫹ Q8 ⫹ Q5 ⫽ QG

H E ⫺ H G ⫽ R 5 # Q5 # |Q5|

Find (H C , H D , H F , H G , Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8,) ⫽

FIGURE F.7

39.862 30.814 30.034 29.717 29.257 0.318 0.135 0.062 0.018 0.044 0.184 0.034 0.072

Example 11.6 Mathcad solution



10:05 PM

Page 766

Appendix

26 8 19 27

Figure 11.7a EPANET 2 solution FIGURE F.8

11 1

1

20

2

2

5

13

12

4

3

28

14

10

15

6

9

3

17

21

16

7

8

18

19

(a) Network map

10

9

22

27

4

15

11

20

25

12

22

21

14

23

24

16

17

5

26

13

6

23

18

7

24

25

766

11/30/10



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Page 767

Appendix 767 Link – Node Table: Link ID 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 1 2 3 4 5 6 7 8 9 10

Start Node 2 3 4 4 3 7 8 8 7 11 12 14 14 12 11 17 17 1 5 6 15 16 13 18 19 10 9

End Node

Length m

3 4 5 6 7 8 9 10 11 12 14 15 16 13 17 18 19 2 20 21 22 23 24 25 26 27 28

195 158 115 155 188 117 59 82 130 102 78 55 60 114 165 64 127 #N/A #N/A #N/A #N/A #N/A #N/A #N/A #N/A #N/A #N/A

Diameter mm 100 50 50 50 100 100 50 50 100 100 100 50 50 50 100 50 50 #N/A Pump 50 Valve 50 Valve 50 Valve 50 Valve 50 Valve 50 Valve 50 Valve 50 Valve 50 Valve

(b) Computed results



768

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Page 768

Appendix

Node Results: Node ID 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 20 21 22 23 24 25 26 27 28

Demand LPS 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -38.91 3.15 3.32 3.13 3.51 3.42 5.36 4.47 5.85 6.70

Head m

Pressure m

135.85 89.79 47.59 40.02 36.36 58.08 53.37 37.67 36.41 49.22 47.26 38.55 46.56 42.98 41.74 46.01 34.75 30.07 30.00 37.00 33.00 40.00 38.00 35.00 26.00 24.00 26.00 24.00

105.85 57.79 12.59 3.02 3.36 28.08 26.37 13.67 10.41 22.22 14.26 3.55 6.56 2.98 3.74 20.01 8.75 6.07 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Quality

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

Reservoir Reservoir Reservoir Reservoir Reservoir Reservoir Reservoir Reservoir Reservoir Reservoir

FIGURE F.8 (b) Computed results (continued )



11/30/10

10:05 PM

Page 769

Appendix 769 Link Results: Link ID

Flow LPS

Velocity m/s

Headloss m/km

Status

11

38.91

4.96

236.24

Open

12

6.47

3.30

267.05

Open

13

3.15

1.61

65.81

Open

14

3.32

1.69

72.49

Open

15

32.44

4.13

168.65

Open

16

12.55

1.60

40.21

Open Open

17

6.70

3.42

266.15

18

5.85

2.98

206.84

Open

19

19.89

2.53

68.13

Open

20

10.06

1.28

19.27

Open

21

6.64

0.85

8.94

Open

22

3.13

1.60

65.00

Open

23

3.51

1.79

80.23

Open

24

3.42

1.74

76.36

Open

25

9.83

1.25

19.45

Open

26

5.36

2.73

176.01

Open Open

27

4.47

2.28

125.52

1

38.91

0.00

⫺105.85

2

3.15

1.61

3.02

3

3.32

1.69

3.36

Active Valve

4

3.13

1.60

2.98

Active Valve

5

3.51

1.79

3.74

Active Valve

6

3.42

1.74

3.55

Active Valve

7

5.36

2.73

8.75

Active Valve

8

4.47

2.28

6.07

Active Valve

9

5.85

2.98

10.41

Active Valve

10

6.70

3.42

13.67

Active Valve

Open Pump Active Valve

(b) Computed results (continued )



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Page 770

Appendix 2

5 5

6 4

6

17

3

7 19 15 1

1

3

2

18

10

4

8

8 14 10 9 9

16

11 15 12

13 11 13 12 14

(a) Network map FIGURE F.9

Figure 11.7b EPANET 2 solution



11/30/10

10:05 PM

Page 771

Appendix 771 Link – Node Table: Link ID 2 3 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 1

Start Node 3 4 5 6 6 7 8 9 9 14 13 11 11 9 10 10 15 1

End Node 4 5 6 2 7 8 9 10 14 13 11 4 12 11 5 15 7 3

Length m

Diameter mm

3000 1520 1520 305 1680 1070 1680 1680 1380 760 1100 2000 1200 670 1520 900 1200 #N/A

600 450 400 150 350 300 350 300 300 150 300 450 400 380 350 350 350 #N/A Pump

(b) Computed results



772

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Page 772

Appendix

Node Results: Node ID 3 4 5 6 7 8 9 10 11 12 13 14 15 1 2

Demand LPS 0.00 0.00 140.00 0.00 100.00 100.00 0.00 140.00 0.00 55.00 55.00 55.00 85.00 -823.11 93.11

Head m

Pressure m

Quality

192.15 155.46 132.55 126.03 122.64 123.04 130.00 124.13 136.56 136.02 133.12 128.61 122.59 15.00 61.00

177.15 109.49 83.55 76.03 73.64 77.04 87.00 80.13 92.56 96.02 92.12 88.61 76.59 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Reservoir 0.00 Reservoir

Link Results: Link ID

Flow LPS

Velocity m/s

Headloss m/km

Status

2

823.11

2.91

12.22

Open

3

460.26

2.90

15.10

Open

4

177.63

1.41

4.29

Open

5

93.11

5.27

213.20

Open

6

84.52

0.88

2.01

Open

8

-22.87

0.32

0.37

Open

9

-122.87

1.28

4.15

Open

10

74.98

1.06

3.50

Open

11

39.19

0.55

1.01

Open

12

-15.81

0.89

5.94

Open

13

-70.81

1.00

3.13

Open

14

-362.85

2.28

9.46

Open

15

55.00

0.44

0.45

Open

16

-237.04

2.09

9.80

Open

17

-142.63

1.48

5.54

Open

18

77.61

0.81

1.71

Open

19

-7.39

0.10

0.05

1

823.11

0.00

-177.15

Open Open Pump

FIGURE F.9 (b) Computed results (continued )


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Bibliography

REFERENCES ADRIAN, R. J., “Multi-point Optical Measurements of Simultaneous Vectors in Unsteady Flow—A Review,” Int. J. Heat Fluid Flow, Vol. 7, No. 2, June 1986, pp. 127–145. BAKHMETEFF, B. A., Hydraulics of Open Channels, McGraw-Hill Book Company, New York, 1932. BEAN, H. S., ed., Fluid Meters: Their Theory and Application, 6th ed., American Society of Mechanical Engineers, New York, 1971. BENEDICT, R. P., Fundamentals of Pipe Flow, John Wiley & Sons, Inc., New York, 1980. CHAPRA, S. C., and CANALE, R. P., Numerical Methods for Engineers, 3rd ed., McGrawHill Book Company, New York, 1998. CHOW, V. T., Open Channel Hydraulics, McGraw-Hill Book Company, New York, 1959. COLEMAN, HUGH W., and STEELE, W. GLENN, JR, Experimentation and Uncertainty Analysis for Engineers, 2nd ed. John Wiley & Sons, New York, 1998. CROSS, H., “Analysis of Flow in Networks of Conduits or Conductors,” University of Illinois Bulletin 286, November 1936. GOLDSTEIN, R. J., ed., Fluid Mechanics Measurements, 2nd ed., Taylor and Francis, Washington, DC, 1996. HEC-RAS River Analysis System, Hydrologic Engineering Center, U.S. Army Corps of Engineers, Davis, CA, July 1995. HENDERSON, F. M., Open Channel Flow, Macmillan Publishing Company, New York, 1966. KARASSIK, I. J., MESSINA, J. P., COOPER, P., HEALD, C. C., eds., Pump Handbook, 3rd ed., McGraw-Hill Professional Publishing, 2000. KING, H. W., and BRATER, E. F., Handbook of Hydraulics, 7th ed., McGraw-Hill Book Company, New York, 1996. KLINE, S. J., ed., Proceedings of the Symposium on Uncertainty Analysis. J. Fluids Eng., Vol. 107, No. 2, June 1985, pp. 153–178. MCBEAN, E. A., and PERKINS, F. E., “Convergence Schemes in Water Profile Computation,” J. Hydraulics Div., ASCE, Vol. 101, No. HY10, October 1975, pp. 1380–1384.

773



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Bibliography MARTIN, C. S., “Experimental Investigation of Column Separation with Rapid Valve Closure,” Proceedings, 4th International Conference on Pressure Surges, BHRA Fluid Engineering, Cranfield, England, 1983, pp. 77–88. ORMSBEE, L. E., and WOOD, D. J., “Hydraulic Design Algorithms for Pipe Networks,” J. Hydraulic Eng., ASCE, Vol. 112, No. 12, December 1986, pp. 1195–1207. ROBERSON, J. A., CASSIDY, J. J., and CHAUDHRY, M. H., Hydraulic Engineering, Houghton Mifflin Company, Boston, MA, 1988. ROBERSON, J. A., and CROWE, C. T., Engineering Fluid Mechanics, 6th ed., John Wiley & Sons, 1997. ROSSMAN, L. A., EPANETZ Users Manual, United States Environmental Protection Agency, June 2000. STEPANOFF, A. J., Centrifugal and Axial Flow Pumps, 2nd ed., John Wiley & Sons, Inc., New York, 1957. SWAMEE, P. K., and JAIN, A. K., “Explicit Equations for Pipe-Flow Problems,” J. Hydraulics Div., ASCE, Vol. 102, No. HY5, May 1976, pp. 657–664. TAYLOR, G. I., “Dispersion of Soluble Matter in Solvent Flowing Slowly Through a Tube,” Proc. Royal Soc. London, Vol. 219A, 1953, pp. 186–203. U. S. DEPARTMENT OF INTERIOR, BUREAU OF RECLAMATION, Design of Small Dams, U.S. Government Printing Office, Washington DC, 1974. VAN DYKE, M., An Album of Fluid Motion, Parabolic Press, Stanford, CA, 1982. WARNICK, C. C., Hydropower Engineering, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1984. WOOD, D. J., “Algorithms for Pipe Network Analysis and Their Reliability,” Research Report 127, Water Resources Research Institute, University of Kentucky, Lexington, KY, 1981. WYLIE, E. B., and STREETER, V. L., Fluid Transients in Systems, Prentice Hall, Englewood Cliffs, N.J., 1993.

GENERAL INTEREST DAILY, J. W., and HARLEMAN, D. R. F., Fluid Dynamics, Addison-Wesley Publishing Company, Inc., Reading, MA, 1968. ESKINAZI, S., Principles of Fluid Mechanics, 2nd ed., Allyn and Bacon, Inc., Needham Heights, MA, 1968. FOX, R. W., and MCDONALD, A. T., Introduction to Fluid Mechanics, 4th ed., John Wiley & Sons, Inc., New York, 1992. FUNG, Y. C., Continuum Mechanics, 2nd ed., Prentice-Hall, Inc., Englewood Cliffs, NJ, 1977. HANSON, A. G., Fluid Mechanics, John Wiley & Sons, Inc., New York, 1967. JOHN, E. A. J., Gas Dynamics, Allyn and Bacon, Inc., Needham Heights, MA, 1969. MUNSON, B. R., YOUNG, D. F., and OKIISHI, T. H., Fundamentals of Fluid Mechanics, 2nd ed., John Wiley & Sons, Inc., New York, 1994. OWCZAREK, J. A., Introduction to Fluid Mechanics, International Textbook Company, Scranton, PA, 1968.



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Bibliography 775 POTTER, M. C., and FOSS, J. F., Fluid Mechanics, Great Lakes Press, Okemos, MI, 1979. ROBERSON, J. A., and CROWE, C. T., Engineering Fluid Mechanics, 6th ed., HoughtonMifflin Company, Boston, MA, 1996. SABERSKY, R. H., ACOSTA, A. J., and HAUPTMAN, E. G., Fluid Flow, 5th ed., Macmillan Publishing Company, New York, 2001. SCHLICHTING, H., Boundary Layer Theory, 8th ed., McGraw-Hill Book Company, New York, 2000. SHAMES., I., Mechanics of Fluids, 3rd ed., McGraw-Hill Book Company, New York, 1992. SHAPIRO, A. H., The Dynamics and Thermodynamics of Compressible Fluid Flow, Ronald Press, New York, 1953. STREETER, V. L., and WYLIE, E. B., Fluid Mechanics, 8th ed., McGraw-Hill Book Company, New York, 1985. VAN WYLEN, G. J., and SONNTAG, R. E., Fundamentals of Classical Thermodynamics, 2nd ed., John Wiley & Sons, Inc., New York, 1976. WHITE, F. M. Fluid Mechanics, 3rd ed., McGraw-Hill Book Company, New York, 1994. YIH, C. S., Fluid Mechanics, West River Press, 1988. YUAN, S. W., Foundations of Fluid Mechanics, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1967.


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Answers Chapter 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.16 1.18 1.20 1.22 1.24 1.26 1.28 1.30 1.32 1.34 1.36 1.38

C B A C C B D A D C C B D c) FL/T e) FT/L a) FT2/L4 b) m/s2 kg/s, N/m, N a) 1.26108 N, c) 6.7108 Pa 2 2 e) 5.210 m a) 5.56105 m/s c) 37.3 kW e) 21010 N/m2 10.06 lb b) 142.2 kPa d) 78.8 kPa 62.8 kPa, 1.95% 50.6°C 2.40 N, 0.0095° 976 kg/m3, 9560 N/m3, 0.2%, 0.36% b) 0.635 kg

1.40 1.42 1.44 1.46 1.48 1.50 1.54 1.56 1.58 1.60 1.62 1.64 1.66 1.68 1.70 1.72 1.74 1.76 1.78 1.80 1.82 1.84 1.86 1.88 1.90 1.92

28.6 N/m3, 6.86106 m2/s 0.007848 N/m2, 0.007848 N/m2 3.2 Pa, 6.4 Pa 2.74 ft-lb, 1.04 hp 91105 ft-lb E(eCy 1) 899 m b) 4940 fps 29.6 kPa, 59.2 kPa 8010 kPa 0.0174 in. 18s/prg 2spD 50°C 7.4 kPa 16.83 km 1.19 kg/m3 9333 N 25.2 m a) 19.25 m/s c) 16.42 m/s 69.2°C 101 Btu b) 243 kJ 804 kPa gage b) 266.9 m/s d) 1301 m/s b) 2854 m

Chapter 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.12

C D C A B A D A A a) 25.5 m

c) 1.874 m

2.14 2.16 2.18 2.20 2.22 2.24 2.26 2.28 2.30 2.32

b) 10.2 m –5.37 psi 103 kPa 96.49 kPa, 96.44 kPa, 0.052% b) 323,200 psf, 322,000 psf, 0.371% a) 69.9 kPa c) 30.6 kPa 33.35 kPa a) 680 kg/m3, gasoline c) 999 kg/m3, water 4.51 psi 17.43 cm

776


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Answers 777

2.34 2.36 2.38 2.40 2.42 2.44 2.46 2.48 2.50 2.52 2.54 2.56 2.60 2.62 2.64 2.66 2.68 2.70 2.72

a) 4580 N/m3 a) 420 kN b) 23.4 kN a) 1.372 m 4.62 in. 8.804 kN 1.23106 people b) 16.4 ft a) 1.089 b) 0.01886 kg gx 8369 N/m3 or gx 1435 N/m3 1.1 GM 0.277 m. stable a) 99.6 kPa c) 17.15 psi b) 4.8 m/s2 a) 640 kN b) 1163 kN b) 9 kPa, 3.11 kPa, 5.89 kPa e) 364 psf, 234 psf, 130 psf 2.104 a) 40.5 kPa, 34.6 kPa, 5.89 kPa c) 947 psf, 817 psf, 130 psf 2.106 a) 8149 Pa c) 17.4 kPa 2.108 a) 6670 N c) 9920 N

15.62 kPa 14.0 kPa 17.89 kPa 5.87 kPa 0.52 kPa a) 3.76 cm 4.51 cm 6934 N 29 400 N vs 28 400 N so it will rise b) 277.1 kN a) 739.7 kN at (0, –0.167) m c) 313.9 kN at (0.9375, –1.5) m 523 kN 3346 N b) 0.6667 m 1.55 m b) 799,000 ft-lb. Will tip 616.1 kN a) FH 706 kN, FV 662 kN 70.07 kN

2.74 2.76 2.78 2.80 2.82 2.84 2.86 2.88 2.90 2.92 2.94 2.96 2.98 2.100 2.102

Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.14 3.16

3.18 3.20 3.22 3.23 3.24 3.26 3.28

D C D C B C B A a) x2 2xy y2 4y. A parabola Eulerian: Several college students would be positioned at each intersection and quantities would be recorded as a function of time. ˆ 13 a) nˆ (2iˆ 3j)/ ˆ ˆ c) nˆ (8i 5j)/ 189 b) 8iˆ 4jˆ d) 2iˆ 114jˆ 15kˆ ˆ a) 40i c) 12iˆ 4kˆ 0 20 0 a) rate-of strain C 20 0 0 S 0 0 0 a) (9.375, 0, 0) m/s2 (0u/0t)i b) 1.875 m/s at t q

3.30 3.32 3.34 3.36 3.38 3.42

3.46 3.48 3.50 3.52 3.54 3.56 3.58 3.60 3.62 3.64 3.66 3.68

9.11 104kg/m3s 0.04 kg/m3s a 0V/ 0t (V # §)V 52 105 jˆ 0.0224 iˆ m/s2 Steady: a, c, e, f, h. Unsteady: b, d, g a) inviscid. c) inviscid. e) viscous inside the boundary layers and separated regions. g) viscous. 11 400. Turbulent a) Re 795. Laminar xT 8.17 m. Laminar u0r/ 0t y0r/ 0y 0 b) 351 fps 57 m/s b) r rc: pT rU2q/2 2 d) u 90 : p90 3rUq /2 2 a) 50r(2/x 1/x ) c) 450r(2/x 1/x2) 9.39 m/s a) 3.14 m/s c) 11.62 fps 12.76 m/s b) 3.516 m/s, 193.6 kPa


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3.70 3.72 3.74 3.76 3.78 3.80

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Answers

19.82 cm a) 6195 Pa c) 5470 Pa a) 37.42 m/s, 71.4 m c) 121.8 fps, 231 ft a) 37.4 m/s c) 118.6 fps b) 43.3 kPa d) 59 Pa Negative pressure, so it will rise

3.82 3.84

High pressure The higher pressure at B will force the fluid toward the lower pressure at A, thereby causing a secondary flow normal to the pipe’s axis. This results in a relatively high loss for an elbow.

Chapter 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.20 4.24 4.26 4.32 4.34 4.36 4.38 4.40 4.42 4.44 4.46 4.48 4.50 4.52 4.54 4.56 4.58

B D A D A C B C D A A C D A A b) Q – W 0 b) Conservation of mass d) Energy equation ˆ j, ˆ 7.07 fps, 0.707 ( iˆ jˆ ), 0.866 ˆi 0.5 j, 8.66 fps, 0 1559 cm3 # dmV/dt m rQ 15 fps, 3.97 slug/sec, 2.045 ft3/sec 0.1509 m3/s 320 m/s, 20.9 kg/s, 4.712 m3/s, 2.512 m3/s b) 4.478 m a) 5 m/s, 320 kg/s, 0.32 m3/s c) 7.5 m/s, 480 kg/s, 0.48 m3/s 5.08 kg/s 0.82 kg/s 4.528 slug/sec, 4.6 slug/sec. rV rV 0.565 m 12.04 m/s 0.244 m/s 27.3 m/s # # rp3(d2 d22)h2/4 h21h1 tan 2f4

4.60 4.62 4.64 4.66 4.68 4.70 4.72 4.74 4.76 4.78 4.80 4.82 4.84 4.86 4.88 4.90 4.92 4.94 4.96 4.98 4.100 4.102 4.104 4.106 4.108 4.110 4.112 4.114 4.116 4.118 4.120 4.122 4.124 4.126 4.128 4.130

3.99104 kg/s b) 8.84 mm/s 14.5 kPa/s 1847 W 0.836°C b) 524 kW 7.93 ft or 1.76 ft 9.94 m/s 60.44 psi 32.1 MPa a) 5.01 m3/s b) 0.485 m3/s 12.35d21d22(H/(d14 d 42))1/2 a) 0.663 m 60.4 m a) 0.131 m 23.5 hp 1304 kW a) 299°C 5390 kW 0.5 W 3820 W, 39.4 kPa, 416 kPa 1.85 m or 2.22 m 0.815 m a) 2.00 # # mfqf FDVq m 3(V22 V21)/2 cy(T2 T1)4 7.37 105 m3/s a) ⬵ 0.32 m3/s b) 679 N d) 127 lb a) 692 N c) 2275 N b) 2280 N, 1071 N 49.6 N 618 kN a) 3.2 m, 8.86 m/s 3.8 ft, 15.8 fps 4420 N


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Answers 779

4.132 4.134 4.136 4.138 4.140 4.142 4.144 4.146 4.148

a) 11.46 kN b) 6.25 kN b) 202 kg/s 147.3 kW 986 kW b) 47°, 29.5°, 3.6 MW a) 11.08 kN, 240 kg/s, 80 kg/s 13.33 m/s 647 hp 110 kN

4.150 4.152 4.154 4.156 4.158 4.160 4.164 4.166 4.168

1440 N, 69.9 m3/s, 62.5% 291 hp, 186.2 slug/sec 141 N 191 ft/sec/ft 3780 N a) 54.9 kW/m 46.9 rad/s 39.6 rad/s 11.73(1 e31.8t)rad/s

Chapter 5 5.2 5.4 5.6 5.8 5.10 5.12 5.14 5.16 5.18 5.20 5.22 5.24 5.26 5.28

See Table 5.1 rdu/dx udr/dx 0 u0r/0x „ 0r/0z 0, 0u/0x 0 „/0z 0 a) yr C/r a) 3y/(0.15x 0.5)2 b) 9y/(0.15x 0.5)4 y const Ay (10 0.4/r 2) sin u (10 80/r3) cos u 0.541 m/s 0.296 m/s a) 0.88 m/s b) 2772 m/s2 2 2 2 3100r/(x y ) 4 (x ˆi y jˆ ) (0p/0u)/r ryryu/r ryr 0yu/0r r(yu /r)0yu /0u

5.30 5.32 5.34 5.36 5.38 5.40 5.44 5.48 5.50 5.52 5.56

r(DV/Dt 2 V ( r) (d /dt) r) §p rg 252y2x9/5 5270y3x13/5, txy 5.01 105 Pa §p rg m§ 2V m§(§ # V)/3 3(V1 V2)y/h4 V1 0p/ 0z m3 0 2yz/0r2 0yz/0r)/r4. 0p/0u m30/ 0r(r2 0yu /0r)4 /r myu /r sin 2u r0u/0t m0 2u/0y2 0.3 10y 苲/Dt K§ 2T rDu rDh/Dt K§ 2T Dp/Dt a) 0u/0t y0 2u/0y2, rc0T/0t K0 2T/0y2 m(0u/0y)2

Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 6.8 6.10 6.12 6.14 6.16 6.18 6.20 6.22 6.24

A A A A C A a) FT/L c) FT2/L4 V /r/m const V 1gH/C FD/r/2V 2 f1(d//, m/r/V) h/d f1(s/gd2, b) s CMy/I V Const.1gH ¢p/rV 2 f1(n/Vd, L/d, e/d) Q/ 2gR5 f1(A/R2, e/R, s)

e) FL

6.26 6.28 6.30 6.32 6.34 6.36 6.38 6.40 6.42 6.44 6.46 6.48 6.50

FD/rV 2d2 f1(m/rVd, e/d, I) T/rV 2D2 f3Re4 FD/rV 2d2 f1(m/rVd, e/d, r/d, Cd2) FL/rV 2/2c f1(c/V, t//c, a) FD/rV 2d2 f1(Vdr/m, d/L, r/rc, V/ d) T/r 2d5 g1(f/ , H/d, //d, h/d) d/D f1(V/Vj, s/rVj2D, m/rVjD, ra/r) m/rVD f1(H/D, //D, gD/V 2) y2/y1 f(gy1/V12) Qm/Qp Vm/2m /Vp/2p, (Fp)m /(Fp)p 2 2 2 3 /m /rpVp2/2p, Tm /Tp rmVm /m/rpVp2/3p rmVm a) 160 kg/s, 15 MPa Full-scale is recommended. 1184 N


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Answers

0.0048 ft, 1.11 nm 6.1 109m2/s. Impossible! a) 0.00632 m3/s b) 12 kN 120 kNm 1633 kW 4.16 N, 95 hp b) 186 m/s, 2080 N a) 6320 rpm b) 2000 rpm

6.68 6.70 6.72 6.74 6.76

5 motions/sec f//V g//U2 0p* 0u* 0 2u* 1 0u* Re b) 2 * * 0t 0x r* 0r* 0r* 1 1 (K/mcp)(m/rU/) Pr Re

Chapter 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.14 7.16 7.18 7.20 7.22 7.24 7.30 7.32 7.34 7.36 7.38 7.40 7.42 7.44 7.46 7.50 7.52 7.54 7.56 7.58 7.60 7.62 7.64 7.66 7.68 7.70

D A D D A B D B A B A C b) 0.0015 m/s Re 700 000. Turbulent a) 12.6 m c) 25.0 m LE 7.2 m. Developed 3.7 m vs 0.72 m a) 20.8 m, 5.2 m 0.123°, 7.79 106 m3/s b) 1.04 104 m3/s. Laminar 1.15 105 m3/s, 1.2 m 0.396 psf 12.1 m/s, 3.5 Pa, 260 m 3.85 106 m3/s a) 0.707r0 b) 0.5r0 0.0188 ft, 0.00163 psf, 217 ft 0.462 m/s, 0.0054 Pa a) 0.0274 Ns/m2 12.1 m3/s, 241 000, 60.4 m/s, 20.1 Pa 0.028 m3/s, 2030 0.0556 cfs a) 18.5 m/s c) 6.15 m/s 32 kPa/m 151 Nm 7.9 Nm 18.9 W 0.040 Nm, 12.6 W, 1650 4pmr12r22L 2/(r22 r12)

7.76 7.78 7.80 7.82 7.84 7.86 7.88 7.90 7.92 7.94 7.96 7.98 7.100 7.102 7.104 7.106 7.108 7.110 7.112 7.114 7.118 7.120 7.122 7.124 7.126 7.128 7.130 7.132 7.134 7.136 7.138 7.140 7.142 7.144 7.146 7.148 7.150

œ œ 2 2 uœ2 51.2 m2/s2, u y 9.7 m /s 2 0.146 ft /s, 0.118, 0.342 in. b) Rough b) 0.262 m/s 8.5, 21.7 fps 40.8 kPa/m 24.2 m/s a) 1.12 m/s, 0.0127 m3/s a) 0.0143 b) 0.0146 a) 0.064 c) 0.034 a) 62 ft c) 42 ft a) 136 kPa c) 20 kPa 11,200 psf 147 kPa b) 0.0033 m3/s b) 0.069 kg/s a) 0.032 m c) 0.031 m a) 0.121 m b) 0.127 m 0.000143 m3/s b) 0.257 m3/s a) 52.6 kPa 11.7 0.0044 m3/s b) 0.011 m3/s 1.3 min. 46.1 m/s Oscillates between laminar and turbulent flow 117 kW, 13.0 m 251 hp, 37.9 ft 0.23 m3/s, 190 kW a) 195 kW b) 81 kPa c) 625 kPa 1.05 MW 0.170 psf a) 1.64 m3/s a) 0.794 m3/s b) 0.86 m3/s 3.91 m 2.04 ft


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Answers 781

Chapter 8 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.10 8.14 8.16 8.18 8.20 8.22 8.24 8.26 8.28 8.30 8.32 8.34 8.36 8.38 8.40 8.42 8.44 8.46 8.48 8.50 8.52 8.54 8.56 8.58 8.60 8.62 8.64 8.66 8.68 8.74 8.76 8.78

C C B B D C C D 3.78 105 m b) 7.9 106. Separated 2659 N, 469 N, 2.63, 0.417 b) 4.58 105 N CD, high/CD,low 2.5 0.5, 1.46 105 a) 1380 N, 25.7 kNm 36 kN a) 27 fps a) 10 hp a) 93.6 N $8800 1.24 hp a) 1.58 m/s c) 2.86 m/s $683 125 km/hr 8.13105 ft, 0.24 – 0.0191 m/s 3.5 hp, 0.12 hp 900 N, 60 N 3.3 N, 0.29 N 4800 N, no cavitation 52 kN b) 1.24 m/s2 0.587 13.8 hp 38 m/s 39.9 m/s a) 7.77% increase a) 5x2 10y2 c) 2x2 y2 100y 50x, 100x 50y 40tan1y/x

8.80

8.82 8.84 8.86 8.88 8.90 8.92 8.94 8.96 8.98 8.100 8.102 8.104 8.108

b) 10y 10 tan1y/x c) 100 50(2/x 1/x2)kPa e) 12.5 m/s2 a) (1 , 0) b) 0.119 c) 1.257 d) 27.5 fps (12, 0), ( 12, 0), 0, 0 a) 0.314 m b) (0, 104) m/s2 c) 13.76 kPa b) 8 sin u c) 58 32 sin2u kPa d) 42.7 kN 100 rad/s, 3660 Pa 472,000 lb (0.729, 0.481) m/s b) 5.04 – a) 3 cm c) 3 cm e) 6 mm 20 200 sin 2(x/2) kPa, 20 sin (x/2) 16/(2.67 x), 256/(2.67 x)3 a) 4.792nx/Uq b) 0.328mUq 2Uq/nx c)

冮

d

0

y cos 3(0.1892Uq/n)y4dy

8.110 6.652nx/Uq, 0.451rU2q Re1/2 , 33%, 36% x 8.112 2.1 cm, 0.001 m, 0.001 m 8.114 a) 1.742nx/Uq, 0.6482nx/Uq, 1.2%, 0.62% 8.116 a) 0.0221 m b) 0.00498 Pa c) 0.299 N d) 0.00391 m/s 8.118 a) 0.0949 m, 0.6 Pa 8.120 a) 0.31 lb 8.122 a) 235 m, 0.0618 Pa b) 0.151 Pa, 585 m 8.124 a) 0.00212 b) 0.229 psf c) 8.09105 ft d) 0.228 ft 8.126 163 kN, 0.89 m 8.132 a) 0.0124 Pa b) 0.0122 m c) 0.00527 m/s d) 0.04 m3/s/m 8.134 a) 32.1 Pa b) 0.0248 m c) 0.109 m3/s/m 8.136 3.85 mm, 0.00127 m/s, 0.011 Pa 8.140 negative, zero, positive, positive, negative

Chapter 9 9.10 9.12 9.14 9.16

0.0069 s 393 m 3.776 s 0.113 fps, 0.021°F

9.18 9.20 9.22 9.24

a) 77.3 m/s a) 0.1473 kg/s b) 0.1393 kg/s a) 0.1472 kg/s b) 0.1423 kg/s 211.3 kPa abs, 7.29 kg/s


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9.26 9.28 9.30 9.32 9.34 9.36 9.38 9.40 9.42 9.44 9.46 9.48 9.50

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Answers

27.83 psia, 0.101 slug/sec, 0.202 slug/sec 97.45 kPa abs, 199.4 kPa abs 168 m/s, 476 m/s 0.1025 slug/sec 494.2 kPa abs, 4.29 kPa abs 0.319 ft, 684 fps, 0.417 ft 129 kPa abs, 1004 m/s 8.16 cm 22.1 cm2, 772°C, 3670 kPa abs 1260 m/s 412 kN a) 3.774 kg/m3, 808 kPa abs, 2.966, 473°C, 0.477 (k 1)/(k – 1)

9.52 9.54 9.56 9.58 9.60 9.62 9.64 9.66 9.68 9.70 9.72 9.74

908 m/s, 1600 kPa abs, 8.33 kg/m3 391 kPa abs, 448°C 102.5 kPa abs, 0.471 kg/s, 0.302 kg/s 49.7 psia, 1020 fps, 1989 fps, 523 fps 6 cm, 9.2 cm 2.39 – a) 1.49, 120.4 kPa abs, 620 m/s 780 m/s 555 kPa abs, 413 kPa abs 39.4°, 867 m/s, 156 °C a) 14.24 kPa abs, 26.4 kPa abs 0.0854, 0.010

Chapter 10 10.2 10.6 10.8 10.12 10.14 10.16 10.18 10.20 10.22 10.26 10.28 10.30 10.32 10.34 10.36 10.38 10.40 10.42 10.44

2.86 m a) 2.15 m, 23.5 m2, 18.6 m c) 4.15 ft, 20.9 ft2, 11.8 ft a) 0.0121 m3/s b) 0.244 m a) –0.17 m c) –0.43 m a) 3.46 m b) 11.51 ft a) 0.3 m a) 5.75 m c) 21.5 ft a) 1.00 m c) 1.81 m a) 1.75 m b) 0.83 m ⬵ 0.95 m a) 16 cfs a) 1.03 ft 28.8 ft M/by2c y2/2y2c yc/y a) 3.87 m, 363 N 7.03 m3/s a) 1.77 m, 0.339 m/s, 3.33 m/s a) 3.43 m3/s b) 0.932 m c) 52 kW 2.55 m, 3.38 kW 11.51 m3/s, 0.346 m

10.46 a) y0 0.094 m, y1 0.316 m, y3 0.088 m, ycj 0.243 m 10.48 y0 2.98 m, M3 followed by hydraulic jump to M2 10.50 M3 followed by hydraulic jump to M1 10.52 yc 1.13 m. ycj 1.35 m. Hydraulic jump occurs upstream, followed by an S1 curve. 10.54 a) yc 1.50 m. A hydraulic jump between A and C. 10.56 a) yc 0.7 m. y0 0.99 m. M3 followed by a hydraulic jump to an M2 10.58 a) 20.2 m b) A2 profile 10.60 15.5 m3/s, S3 10.62 Q 7.72 m3/s, yc 0.72 m, y0 1.17 m, ycj 0.41 m. M1 upstream, M3 downstream 10.64 S3 followed by hydraulic jump to S1 10.66 a) 0.87 m b) 48 m 10.68 a) 721 kW b) 59.3 m 10.70 At y 2 m, x ⬵ 55 km 10.72 3.2 m

Chapter 11 11.2 11.4 11.6 11.8 11.10 11.12 11.14

a) 750 W a) 1.29 ft a) 0.265 m3/s a) 0.32 m3/s 1.78 m3/s, 0.285 m3/s, 0.935 m3/s, 1.07 MW 0.198 m3/s, 0.138 m3/s, 0.06 m3/s 16 L/s, 7.1 L/s, 5.1 L/s, 3.8 L/s

11.16 11.18 11.20 11.22

c) 28.2 L/s, 10.7 L/s, 8.6 L/s, 9.1 L/s a) 143 L/s, 290 L/s, 166 L/s 1.05 m3/s, 90.8 m, 73.6 m, 810 kW a) 2.27 m3/s, 0.811 m3/s, 0.580 m3/s, 0.880 m3/s 11.24 a) 2.28 ft c) 3600 gal/hr 11.26 a) 7400 gal/hr


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11.28 2.30, 0.43, 2.73 flow units 11.30 a) 96, 92, 88, 82 m b) 843, 706, 813, 804 kPa 11.32 21.6 L/s, 6.6 L/s, 28.4 L/s 11.34 2.77 m3/s

11.36 11.38 11.44 11.46 11.48

9.3, 1.6, 7.7, 0.6, 7.1 L/s d) 2.06 m3/s 69 s 1.57 L/s, 8.03 s a) 1090 m/s b) 590 kPa

Chapter 12 12.2 12.4 12.6 12.8 12.12 12.14 12.16 12.18 12.20 12.22 12.24 12.26 12.28

42.1 L/s, 35.6 N, 2980 W, 7.22 m 54.1 ft, 1.31 hp 1.23 m 17.6 m, 172 kW P 1.30, so mixed-flow pump a) 2.27 ft, 17.8 ft. 22.7 ft, 113 hp P 0.751, so radial-flow pump Speed and diameter are 1.19 times greater Mixed flow, 0.251 m, 282 rad/s 2.75 m3/min, 12.6 m, 7.2 kW 2.44 m3/min, 10.0 m, 5.1 kW P 5.19 so axial flow a) 280 m3/h, 64 kW, 8.3 m

12.30 12.32 12.34 12.36 12.38 12.40 12.42 12.44 12.46 12.48 12.50

Three pumps, 206 hp a) 5.67 m b) 15.8 m a) radial flow, 9 cm, 4680 rpm b) 3.95 m 6.1°, 28.3 MNm, 243 m, 360 MW 0.736 ft, 399 rpm 189 kW, Francis, 2920 rpm 4.78 m3/s, 1.95 m, 244 mm, 3 jets, 0.204 25.8 ft, Francis or pump/turbine a) 1.45 m b) 13.6 cm 9.2 MW total, two units a) 24.6 kW b) T 1.65, Francis c) 0.3 m, 680 rpm, 26 kW

Chapter 13 13.2 13.4 13.6

a) 46.8 b) 51.6 1.114 104 ft3/sec, 2.16 104 slug/sec, 2.04 fps, laminar 0.0592 m3/s, 7.54 m/s

13.8 a) 0.064 m3/s 13.10 a) 0.0974 m3/s, 13.14 0.00396, 0.50

c) 3.33 cfs





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Index A Absolute pressure, 11 Acceleration, 91–94 in Cartesian, cylindrical and spherical coordinates, 96 component, 209 convective, 93 local, 93 of reference frame, 94 Acoustic flowmeter, 674–675 Acoustic perturbations, 659–660 Actual flow boundary layer, 385 Adams-Bashforth formulas, 711 Adams-Mouton formulas, 711 Added mass, 366–367 exercises dealing with, 415 Adiabatic, 28 Adiabatic process, 427 Adverse pressure gradient, 351 Air properties, 744 Air flows over wedge example, 460 Airfoils flaps, 368 inviscid flow, 102 lift and drag, 367–372 lift and drag coefficients, 369 swept-back, 370 Album of Fluid Motion, 679 Alternate depths, 487 Altitudes, 45 Amplification factor, 716 Analytic function, 374 Andrade’s equation, 17 Anemometers, 663–664 Angular velocity, 94–99 Angular acceleration, 94 Apparent shear stress, 295 Archimedes, 61 Archimedes’ principle, 61 Areas

properties, 743 Atmosphere properties, 739 Atmospheric pressure, 45 Averaging measurement, 659 Axial-flow pump, 603, 611 performance curves, 614, 622 Axial-flow turbine, 640 Kaplan type, 680

B Baffle blocks stilling basin, 508 Bernoulli, Daniel, 108 Bernoulli’s equation, 107–116, 149, 240, 382–383, 495, 662, 667 example, 115 exercises dealing with, 121–125 Bias, 684 Bingham fluids, 17 Blades, 169 Blasius, Paul R. H., 393 Blasius formula, 393 Blunt body vs. streamlined body, 348 Boiling, 22 Borden gage, 660 Boundary conditions, 205 gravity effects, 258–259 Boundary layer, 102 actual flow, 385 air, 387 airfoil, 367 control volume, 388 curved surface, 385 exercises dealing with, 419–420 flow flat plate, 105, 390, 403 separation, 351 theory, 385–409 with transition, 386 transition on separation, 353 viscous effects, 387

Branch piping, 558–562 British Gravitational System, 5 Broad-crested weir, 496 Bubble, 19–20 Buckingham, Edgar, 242 Buckingham p-theorem, 238, 241–245, 261 Buffer zone, 299, 399 Bulk modulus, 429 of elasticity, 18–19, 581 Buoyancy, 61

C Cameras flow visualization, 678 Capillary tube, 20 rise of liquid, 245, 21 Casing, 602 Cauchy, Augustin L., 374 Cauchy-Riemann equations, 374 Cavitating propeller, 23 Cavitation, 22, 111, 363–366, 627, 640 turbomachines, 615–619 Cavitation bubble distribution, 617 Cavitation number critical, 364, 618 turbines, 640–641 Celerity, 517 Center of buoyancy, 64 Center of pressure, 53 Centrifugal pump, 603 Centroids, 52, 53 Channel constriction, 491–492 Channel entrance critical flow, 493 Channel flow over obstacle, 500 Channel geometry, 480–482 Characteristic curves pumps operating in parallel, 632 pumps operating in series, 633 Chemical indicators, 676–677 Chezy-Manning equation, 483, 485, 519, 521, 549 comparisons, 550

785


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Index

Choked flow, 436, 491 Choking condition, 491 Chord, 350 Circle areas, 743 Circular cross section, 481 Circulation, 378 Coefficient of compressibility, 19 Colebrook comparisons, 550 Colebrook equation, 307 Colebrook formula, 550 Common dimensionless parameters, 246–247 Completely turbulent regime, 307 equation, 307 Complex velocity potential, 374 Components, 546 Compressibility, 18–19 exercises dealing with, 34 Compressible flows, 107, 256, 426–467 in elastic pipe, 578–584 tables for air, 744–752 Computational fluid dynamics (CFD), 698–732 described, 700 Condensation shock, 455 Condenser microphone, 661 Cone volumes, 743 Confined flows, 250 Confined incompressible flow, 251 Conical expansion loss coefficients, 315 Conjugate depths, 502 Conservation laws, 23–24 Conservation of energy, 24, 129 Conservation of mass, 23, 129, 137–142 exercises dealing with, 182–187 Conservation of momentum, 24, 129 Conservative property FD, 718 Consistency FDE, 713 Constant density, 106 Constitutive equations, 216 Continuity equation, 138, 214, 226, 754 differential form, 206 Mach numbers, 444–449 Continuum, 10 Contracted rectangular weir, 497 Control location, 499 Controls, 517 Control surface, 132 Control volume, 131 example, 139–143 inlet area, 137 system, 133 time derivative, 135

Convective acceleration, 93 Convergence FD, 717 Converging-diverging nozzle, 436–437 example, 440–441 illustration, 450 shock waves, 449–453 Converging nozzle, 436 Coriolis acceleration, 94 Coriolis-acceleration flowmeter, 675 Correlation coefficient normalized turbulent shear stress, 297 Couette, 282 Couette flow, 284 numerical solution, 719 Crank-Nicholson formula, 711 Crank-Nicholson method, 708 Creeping flows, 346 Critical cavitation number, 618 hydrofoil, 365 Critical depth, 487 Critical energy, 487 Critical flow, 517 channel entrance, 493 Critical Reynolds number, 104, 387 flat plate, 105 Critical zone, 307 Cross-flow turbine, 648 Cup anemometer, 663 Current meter, 663 Curved surface boundary layer, 385 Cylinder added mass, 366 symmetrical flow example, 162 volumes, 743

D Darcy, Henri P.G., 278 Darcy-Weisbach equation, 278, 309, 325, 548 Data digital recording, 683–684 Deflection angle, 457 Deflectors, 164–175 flow example, 167, 168 Del, 207 Density, 4, 10, 14–15 exercises dealing with, 33 variations flow, 107 Derive difference operators, 703 Derived units, 6 Design discharge, 630 Design head, 630 Detached shock waves, 460 Developed flow, 101, 272–274 in circular pipe, 275–280 exercises dealing with, 330 between parallel plates, 281–286

Developed laminar flow, 272 Developed pipe flow losses, 306–312 Developed turbulent flow categories, 309 Difference operator, 705 derivation steps, 703–705 select, 706 Difference operators choice, 710–712 Differential continuity equation, 205–210 exercises dealing with, 231–233 Differential energy equation, 223–228 exercises dealing with, 235 Differential equations, 294 laminar and turbulent flows, 229 Differential forms deriving, 204 fundamental laws, 203–230 Differential momentum equation, 210–222 exercises dealing with, 233–234 Differential pressure meters, 666 Diffuser, 433, 637 purpose, 438 Digital recording data, 683–684 Dilatants, 17 Dimension, 4–8 exercises dealing with, 31 of quantities, 240 Dimensional analysis, 237–262 exercises dealing with, 263–265, 651–652 shedding frequency, 359 Dimensional homogeneity, 238 Dimensional similitude, 619 exercises dealing with, 651–652 Dimensionless coefficients turbomachine, 619–623 Dimensionless form differential equations, 258 Dimensionless parameters, 239–240, 243 Dimensionless performance curves, 622 Dimensionless pressure drop vs. dimensionless velocity, 241 Dimensionless velocity vs. dimensionless pressure drop, 241 Direct integration methods, 528–530 Discharge, 138, 552–553 coefficient, 495 Discretization of domain, 700 of governing equations, 702–705 Displacement thickness, 389 Dissipation function, 226 Dissipative property FD, 719 Divergence of velocity, 207 Double-suction impellers, 602


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Index 787 Doublet, 379 strength, 378 Draft tube, 635 Drag, 348–350 airfoil, 367 exercises dealing with, 415–416 Drag coefficients, 349, 508 around immersed bodies, 352–358 blunt objects, 356 of finite-length circular cylinders, 355 flow around cylinder and sphere, 354 hydrofoil, 364 of infinite-length elliptic cylinders, 355 Mach number, 370 vs. Reynolds number, 254 zero cavitation number blunt objects, 364 Drag force example, 156 Droplets, 19–20 Dual-beam laser-doppler velocimeter, 665 Dynamic response measurement, 659 Dynamic similarity, 248

E Eddy viscosity, 297 Edge of boundary layer, 102, 385 Efficiency losses, 148 of pumps, 630 to size, 622 Elbow flow, 315 Elbow meter, 671–672 Electromagnetic flowmeter, 673–674 Elements, 546 Elemental approach, 275 Ellipse areas, 743 Energy concepts exercises dealing with, 532–535 in open-channel flow, 486–489 Energy correction factor, 151 Energy equation, 144–156 application, 149 Bernoulli’s equation, 115 control volume, 318 differential, 223–228, 235 exercises dealing with, 188–194 Mach numbers, 444–449 steady uniform flow, 149 in transitions, 490–495 Energy grade lines (EGL), 319–322 Energy losses in expansions and contractions, 495 English units, 5 Enthalpy, 27 definition, 210 Entrance flow, 272–274

exercises dealing with, 330 Entrance length, 272 Entrance region of laminar flow, 272–273 of turbulent flow, 273 Entropy, 427, 445 Euler, Leonhard, 89 Eulerian descriptions, 132–136 of motion, 88–89 Euler implicit scheme, 707 Euler number, 246, 248–249 Euler’s equations, 213–215 inviscid flow, 224 Euler’s turbomachine relation, 605 Exit area position, 136 Exit flow, 112 Exit pressures, 441 Exosphere, 44 Expansion waves, 461–466 exercises dealing with, 471 Explicit method spatial derivatives, 706 Extensive property, 24 illustration, 132 rate of change, 135 system, 131 E-y curve, 487–488

F Favorable pressure gradient, 351 Ficticious velocity, 397 Finite-difference methods (FDMs), 700–730 First law of thermodynamics, 24, 25, 129–130 exercises dealing with, 36 Fixed cavitation, 363 Fixed control volume example, 131 system, 133 Fixed-grade nodes, 563 Flapped airfoil, 369 Flat plate critical Reynolds number, 105 laminar portion, 395 Flow. See also Compressible flows; Developed flow; Free-surface flow; Gradually varied flow (GVF); Incompressible flow; Inviscid flow; Isentropic flow; Laminar flow; Steady flow; Turbulent flow; Uniform flow actual boundary layer, 385 air over wedge example, 461 airfoil, 368 around blunt object, 352, 460 around circular cylinder, 354, 382 around immersed bodies, 352–367 around rotating cylinder, 383

around sphere, 110, 354 around wedge, 460 channel over obstacle, 500 choked, 436, 491 between concentric cylinders, 288 confined, 250 confined incompressible, 251 Couette, 284 creeping, 346 critical, 493, 517 density variations, 107 in elbow, 315 entrance, 272–274, 330 equations, 372–377 exit, 112 external, 102 field, 89 exercises dealing with, 117–119 fluid, 100–107, 119–120, 238 free-stream, 349 function varied, 788–759 gas, 442 gravity branch piping system, 559 immersed examples, 347 incompressible 106, 207 intermittent, 105 internal, 271–329 mass rate, 138 measurements, 495–499, 666–675 methods, 499 uncertainty analysis, 684–690 nonhomogeneous, 207 nonuniform, 477, 512 normal-shock, 746–750 nozzle, 670, 671 flow coefficient K versus Reynolds number, 670 one-dimensional, 101, 281, 477 open channels, 475–529 energy concepts, 486–489 momentum concepts, 500–511 with outer cylinder fixed, 290–292 past a circular cylinder, 346 patterns in turbopumps, 604 periodic, 257 in piping systems, 545–585 minor losses, 314–319 quantities, 277–281 plane, 100, 112, 220 Poiseuille, 275, 277 pump-driven branch piping system, 559 rapidly varied, 478 rate, 138 turbomachine coefficient, 620 wrought iron pipe, 318


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Index

Flow (Continued) ratios of forces, 247 Reynolds number high, 254–255, 346, 347, 359 low, 346 separated, 147 slider valve, 240 Stokes, 346 stratified, 207 subsonic, 430 supersonic, 430 around convex corner, 462 large turning angles, 464 three-dimensional, 100, 112 two-dimensional, 100 unsteady, 477, 576–584 vapor, 455–457, 472 viscous, 101 visualization, 675–683 vortex, 383 work, 146 Fluid flows, 238 classification, 100–107 exercises dealing with, 119–120 Fluid mechanics common dimensionless parameters, 248 measurements, 657–695 exercises dealing with, 695–697 symbols and dimensions of quantities, 242 Fluid motion description, 88–99 Fluid particle, 295 infinitesimal parallelepiped, 95 Fluid properties, 14, 737–742 Fluid statics defined, 40 Fluid velocity measurement, 661–666 Forces, 5 on bend, 150 on curved surfaces, 57–60 exercises dealing with, 80–83 on floating object, 62 on nozzle, 157 on plane areas, 51–56 exercises dealing with, 77–80 plane surface, 53 on rectangular gate, 53 on submerged body, 61 vector, 9 Fourier, Jean B. J., 223 Fourier’s law of heat transfer, 223, 704 Fourier method, 715 Four-stage centrifugal pump, 634 Francis spiral tubing, 636 Francis turbine, 635 isoefficiency curve, 638 performance curve, 639 Free-body diagram, 56–60

Free stream, 105 flow, 349 fluctuation intensity, 351 Free surface, 43, 70, 250, 476 flow, 250–253, 476 classification, 477–478 force on gate, 159 Frictional losses minor losses, 314 in pipe elements, 547–551 Friction factor, 278, 306 uniform flow, 325 Froude, William, 246 Froude number, 246, 248–249, 257, 261, 489 free-surface flow, 252–253 significance, 478–479 Fundamental dimensions and their units, 6 Fundamental laws differential forms, 203–230 integral forms, 180

G Gage pressure, 12 Gaging site, 499 Galloping, 361 Gas, 9 continuum view, 8–11 flows, 442 Gauss-Legendre formula, 526 Gauss-Legendre quadrature, 526, 527 Gauss-Seidel method, 724 Gauss’s theorem, 204 General energy equation, 146 Generalized cross-section, 489–490 geometry, 481–482 Generalized network equations, 565–567 Geometric similarity, 249 Gird-independent solution FD, 719 Gradient of scalar function, 258 Gradient operator, 207 Gradually varied flow (GVF), 478 differential equation, 512–514 notation for computing, 521–522 Gravity effects boundary condition, 258–259 flow branch piping system, 559 flow pattern, 250

H Hardy Cross method, 552, 568–572 Hazen-Williams coefficient, 549 nominal values, 550 comparisons, 550 equation, 548

Head-discharge relations real fluid flow, 610 Head loss, 148, 278, 285, 306 coefficients, 316 sudden expansion, 163 Heat transfer, 144 HEC-RAS River Analysis System web site, 528 Helium bubbles, 677 Hemisphere volumes, 743 High Reynolds numbers, 254–255, 346, 347, 350, 359 Holography, 680 Homogeneous fluid, 216 Horizontal pipe bend flow example, 160 Horizontal rotation, 69 Hot wire anemometer, 664 Hydraulically smooth, 299 Hydraulic bore, 506 Hydraulic depth, 489 Hydraulic grade lines (HGL), 319–322 Hydraulic jump,161, 491, 503–510, 519 horizontal rectangular channels, 505 translating, 506 Hydraulic radius, 481 Hydraulic turbines applications ranges, 647 Hydrodynamic perturbations, 659–660 Hydrofoil, 364 Hydrogen bubbles, 676 Hydrometer, 63 Hydrostatic pressure distribution, 479–480 Hypersonic flow, 430

I Ideal gas assumption, 226 exercises dealing with, 36 properties, 740 Ideal gas law, 427 Idealized axial-flow impeller, 613 Idealized Francis turbine runner, 637 Idealized hydraulic jump, 504 Idealized radial-flow impeller, 604 Ideal plastics, 17 Infinitesimal control volume, 206 Immersed flows examples, 347 Impact tube, 662 Impellers, 602 pump types, 603 Implicit method spatial derivatives, 706–707 Impulse turbine, 634, 641–646 Pelton type, 642


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Index 789 Incompressible flow, 106, 207, 214 in inelastic pipe, 576–577 Reynolds number, 250–253 Incompressible fluids study of, 26–27 Inertial moment, 176 Inertial reference frame, 94 Infinitesimal control volume, 206 Infinitesimal fluid particle, 212 Infinitesimal parallelepiped fluid particle, 95 Initial conditions, 205 Initial-value problems (IVPs), 711 Inlet area position, 136 Inner region, 397 Integral forms of fundamental laws, 180 Integrands distributions, 204 Intensive property, 24 Interfaces, 250 Interferometry, 681–683 Intermittent flow, 105 Internal flows, 271–329 Internal friction, 147 Internal inviscid flows, 110 International System, 5 Inviscid core length, 272 Inviscid flow, 101, 105, 111, 347 airfoil field, 367 boundary layer, 385 internal, 110 Ionosphere, 44 Irregular channels, 528 Irregular section, 480 Irrotational flow, 94, 372 Irrotational vortex, 379 Isentropic, 427 expansion vapor, 454 waves, 461–464 flow, 744–746 exercises dealing with, 37, 467–469 nozzle, 431–442 process, 28 Isoefficiency curve Francis turbine, 638 Pelton turbine, 644 Isothermal atmosphere, 47 Isotropic fluid, 216 Isotropic Newtonian fluids, 239–240 Iterative procedure discharges, 556 hydraulic grade line, 556

J Jain formula, 309 Joukowsky equation, 580

K Kaplan type axial-flow turbine, 640 Karman integral equation, 389 Karman vortex street, 359 Kinematic similarity, 249 Kinematic viscosity, 297 example, 551 Kinetic-energy correction factor, 151 example, 155 Kinetic pressure, 258 Kinetic viscosity as function of temperature, 805 Knock, 361 Kutta-Joukowsky theorem, 384

L Lagrange, Joseph L., 89 Lagrangian descriptions, 132–136 of motion, 88–89 Lagrangian-to-Eulerian transformation, 135 Laminar boundary layer approximate solution, 389–393 equations, 402 exercises dealing with, 422–423 exercises dealing with, 421–422 on separation, 353 solution for, 405 Laminar flow, 102 differential equations, 229 equation, 286 exercises dealing with, 330 between parallel plates, 281–287 exercises dealing with, 332–334 in pipe, 274–281 exercises dealing with, 331–332 Reynolds number, 288 rotating cylinders, 104 between rotating cylinders, 288 exercises dealing with, 334–335 velocity, 103 Laminar portion flat plate, 395 Laminar profile, 387 Laplace, Pierre S., 373 Laplace’s equation, 373, 377–380 Laser-doppler velocimeter (LDV), 665 Laser induced Fluorescence (LIF), 680–681 Leibnitz’s rule, 502 Lift, 348–350 airfoil, 367 coefficient, 349 hydrofoil, 365 Lift and drag on airfoils exercises dealing with, 415–416 Lighting, 678 Linearization of system energy equations, 567–568

Linearly accelerating containers, 67–69 exercises dealing with, 84 Linearly accelerating tanks, 68 Line source, 379 Liquid-gas surfaces, 250 Liquids, 9 continuum view, 8–11 properties, 740 Local acceleration, 93 Local flow parameters measurement, 658–666 Local Reynolds number, 387 Local skin friction coefficient, 391, 399, 405 Loss coefficients, 148 conical expansion, 315 fitting, 316 Losses, 147 developed pipe flow, 306–319 efficiency, 148 noncircular conduits, 313 pipe flow, 314–319 in piping systems, 339–342, 546–551 Low-Reynolds-number flows, 346

M Mach, Ernst, 106 Mach number, 106, 246, 248–249, 257, 369, 427–431 drag coefficient, 370 Mach waves, 430, 463–465 Mach-Zehnder interferometer, 683 Manning n, 326 Manometers, 48–51, 659 exercises dealing with, 73–74 Mass flux, 138, 205 Mass rate of flow, 138 Material derivative, 93 Mathcad, 689–691, 693–694 example, 527 measurements in fluid mechanics, 688 Mean free path, 10 Metacentric height GM defined, 64 Metastable equilibrium, 455 Method of least squares measurement in fluid mechanics, 690–693 Microhydro, 648 installations, 647 Micromanometer, 48, 51 Minihydro, 648 installations, 647 Minor losses pipe flow, 314–319 in pipes and conduits, 339–342 Mixed-flow pumps, 603, 611–615 performance curves, 616, 623 Mixing length, 297 Mole, 5


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Index

Molecular diffusion, 720 Molecular Tagging Velocimetry (MTV), 677 illustrated, 680 Moment-of-momentum equation, 130–132, 176–178 Moments, 53 Momentum and energy exercises dealing with, 200–201 Momentum concepts exercises dealing with, 535–537 in open-channel flow, 500–511 Momentum-correction factor, 172–173 Momentum equation, 157–175, 500–502 differential form, 210 exercises dealing with, 195–200 Mach numbers, 444–449 numerical solution, 510–511 transition region, 502–503 Momentum thickness, 389 Moody, Lewis F., 306 Moody diagram, 306, 308 Motivation, 239–240 Multistage pumps, 634

N Navier, Louis M. H., 217 Navier-Stokes equation, 216–218, 258–259, 260, 275, 282, 289–290, 294, 402, 700 airfoil, 367 integrating, 283–284 solving, 276, 283, 289 Negative pressure gradient, 407 Net positive suction head (NPSH), 617–618 curves, 611 turbines, 640–641 Network analysis using computer software, 573–576 Newtonian fluids, 17, 216 isotropic, 239–240 Newton’s second law, 5, 7, 24, 107, 130, 157–175, 204, 241, 274 applied to fluid particle, 211 Noncircular conduits losses, 313 Nonhomogeneous flow, 207 Noninertial reference frames momentum equation, 174–175 motion relative to, 94 Non-isotropic fluids, 239–240 Non-Newtonian fluid, 17 Nonuniform flow, 151, 477 Nonuniform gradually varied flow, 512 exercises dealing with, 537–543 Nonuniform velocity profiles, 138 Normal depth, 483 Normalized differential equations, 258–261 exercises dealing with, 268–269

Normalized problem similitude conditions, 261 Normal shock in air entropy change, 445 exercises dealing with, 469–470 flow, 746–751 wave, 442–454 T-s diagram, 447 Normal stress, 9, 211 Normal vector, 132 No-slip condition, 17 Nozzle, 433. See also Converging nozzle isentropic flow, 431–442 purpose, 437–438 supersonic, 434, 438 venturi, 111 Numerical errors FD, 718 Numerical integration method, 526 Numerical stability FD, 714–717

O Oblique shock wave, 456–460 exercises dealing with, 470 Oil films, 678 One-dimensional flow, 100, 281 One-dimensional free-surface flows combinations, 477 Open channel flow, 477–480 energy concepts, 486–489 exercises dealing with, 343 momentum concepts, 500–511 flowmeters, 675 gate energy equation, 149 Open rectangular channel flow example, 161 Operating point, 323, 630 Ordinary differential equations (ODEs), 711 Orifice meter, 667, 669–671 Outer region, 299, 301, 397

P Parabolic profile example, 173 Paraboloid of revolution, 70 Parallel piping, 555–558 Parallel plates stationary and moving flow, 701 Partial differential equations (PDEs), 700–701 Particle image velocimetry (PIV), 677 illustrated, 679 Particles in air, 677

Particle velocity measurement, 662 Pathline, 90 Pelton bucket velocity vector diagram, 642 Pelton turbine isoefficiency curve, 644 performance curve, 645 speed factor versus efficiency, 644 Pelton type impulse turbine, 642 Pelton wheel, 639 Performance curves axial-flow pump dimensionless, 621 vane angle comparison, 614 Francis turbine, 639 mixed-flow pump dimensionless, 621 impeller and vane angle comparison, 616 Pelton turbine, 643, 645 radial-flow pump dimensionless, 620 impeller comparison, 612 real fluid flow, 610 Periodic flows, 257 Persistence of irrotationality, 220 Photography, 678 Physical quantities, 4–8 Piezoelectric pressure transducer, 661 Piezometer, 109 opening example, 115 Piezometric head, 43, 148, 552–553 Pilot probe, 109 Pipe bend horizontal flows example, 160 Pipe elements frictional losses, 547–551 Pipe equation smooth, 307 Pipe flow developed losses, 306–312 minor losses, 314–319 quantities, 277–281 smooth exponent, 302 Pipe networks analysis, 563–576 Piping systems, 547, 552–562 exercises dealing with, 340–342 flows, 545–585 losses, 546–551 with pump, 323 Pitot probe, 109 example, 453–454 Pitot-static probe, 109 measurement, 662


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Index 791 Plane flow, 100, 112, 220 Poiseuille, Jean L., 275, 277 Poiseuille flow, 276, 277 Positive pressure gradient, 407 Positive surge waves, 506 Potential flow numerical solution, 723 theory, 372–384 Power, 291 coefficient turbomachine, 620 law form, 393–396 law profile, 301, 302 output, 164 to rotate a cylinder, 291 Prandtl, Ludwig, 403 Prandtl boundary layer equation, 403 Prandtl-Meyer function, 463, 749–750 Prandtl tube, 662, 665 Precision, 685 Pressure in atmosphere, 44–47 coefficient turbomachine, 620 defined, 40 drop, 306 vs. velocity curves, 240 exercises dealing with, 32, 73–74 in fluid, 41 force, 20 gradient, 406–409 exercises dealing with, 423 influence of, 408 head, 148 in liquids at rest, 43–44 measurement, 659 and motion, 40 at a point, 40–41 prism, 54 probes, 109 pulse wave speed, 581 recovery factor, 438 scales, 11–13 transducer, 660–661 variation, 41–43 in horizontal pipe flow, 275 Profile development region, 272 Profile synthesis, 518–520 example, 519 Propeller anemometer, 663 fluid flow, 170 momentum equation, 170 velocity, 171 Prototype velocity wind-tunnel airspeed, 256–257 Pseudoplastics, 17 Pulsed light velocimeter, 677–678 illustrated, 678

Pump characteristic curves, 194, 323 and system demand curve, 629 Pump-driven flow branch piping system, 559 Pump head, 150 discharge curve, 629 Pump housing, 602 Pump impellers types, 603 Pumps matching to system demand, 629–630 multistage, 634 in parallel and in series, 630–633

Q Quantity means of measurement, 666 Quasi-equilibrium process, 28 Quasi-static process, 28

R Radial-flow pumps, 602, 603–611 efficiency function of specific speed and discharge, 627 performance curves, 611, 612, 622 Rake, 666 Rankine-Hugoniot, 469 Rapidly varied flow (RVF), 478 Rate meters, 666 Rate-of-energy transfer control surface temperature difference, 144 Rate-of-heat transfer, 144 fluid element, 223 Rate-of-strain tensor, 96 Ratio of specific heats, 28 Rayleigh-pitot-tube formula, 454 Reaches, 546 turbines, 634–641 Francis type, 636 Rectangle areas, 743 section, 481, 487 sharp-crested weir, 497 Refraction methods index, 681–683 Regression analysis measurement in fluid mechanics, 690–693 Regular section, 480 Relative roughness, 299 Repeatability, 685 Repeating variables, 243 Representative controls, 518 Resistance coefficient, 548 Resultant force, 57 Reversible pump/turbine units, 648 Reynolds, Osborne, 104 Reynolds numbers, 104, 246–249, 257, 260, 286, 549 drag coefficients, 254, 355

example, 310–312 high, 254–255, 346, 347, 350, 359 incompressible flow, 250–253 laminar flow, 273 local, 387 low, 346, 362 separation, 352 turbomachine, 618 uniform flow, 325 Reynolds stress, 659 Reynolds transport theorem, 135 Rheology, 9 Riemann, Georg F., 374 Rocket, 175 Rotameter, 677 Rotating arms,177 Rotating containers, 69–71 exercises dealing with, 85 Rotating cylinders laminar flow, 104, 288–292 Rough pipe, 301 Runge-Kutta formulas, 712 Runner, 634

S Scalar potential function, 375 Schlieren, 681–683 Second coefficient of viscosity, 216 Self-similar relation, 397 Semicircle areas, 743 Semicircular arc streamlines, 104 Semiellipse areas, 743 Separated flows, 147 exercises dealing with, 410 Separated region, 111, 347 Separation, 350–352 Sequent depths, 502 Series piping, 552 Shadowgraph, 681–683 Shaft work, 146 Sharp-crested weir, 496 Shear stress, 9, 107, 211, 223, 287, 296 components, 213 Shear velocity, 299, 397 Shear work, 146 Shock waves, 430, 442–449 in converging-diverging nozzles, 449–453 oblique, 456–461 Shroud, 602 Similarity rules family of geometrically similar turbomachines, 623–625 Similitude, 237–262 conditions normalized problem, 261 defined, 238 exercises dealing with, 263–265


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Index

Simplified flow situation, 284–287 Simpson’s rule methods, 526 Single-suction pump, 603 SI (Systeme International) prefixes, 7 units, 5 Skin friction coefficient, 391, 405 Slider valve flow, 240 Slots, 368–370 Smooth pipe equation, 307 exponent, 302 Smooth wall, 299 Solution algorithm defined, 762 Sound wave, 428 Source strength, 378, 381 Spatial derivatives explicit method, 706 implicit method, 706 Specific discharge, 487 Specific energy, 130, 486–490 Specific gravity, 14–15 Specific speed, 626–628 Specific weight, 14–15 exercises dealing with, 33 Speed factor versus efficiency Pelton turbine, 645 Speed of sound, 427–431 exercises dealing with, 37, 466–467 Sphere added mass, 366 volumes, 743 Sprinkler, 177 Stability, 64–67 exercises dealing with, 83 of floating body, 65 numerical, of submerged body, 65 Stage-discharge curve, 499 Stagnation point, 100, 383 Stagnation pressure, 109 Stagnation quantities, 434 Stall, 111, 349, 350 Standard atmosphere, 44–46 conditions, 10 Standard step method, 521–525 Static pressure, 109 head example, 114 Static tube, 663 Stationary deflector, 164 Stationary shock wave, 444 Steady flow, 89, 91, 180, 477 exercises dealing with, 585–598 Steady nonuniform flow, 151–156, 180 in channel, 478 momentum equation, 172–173 Steady uniform flow, 148–150 incompressible, 180 momentum equation, 158–170

Stilling basin, 508 Stokes, George, 217 Stokes flows, 346 Strain-gage transducer, 661 Strain rate, 15 Stratified flow, 207 Stratosphere, 44, 46 Streakline, 90 Stream function, 374 exercises dealing with, 416–417 Streamline, 91, 362–363 body vs. blunt body, 348 exercises dealing with, 414–415 pattern, 249 pressure, 111 semicircular arc, 104 Stream lining, 362 Streamtube, 91 Streamwise-component momentum equation, 428 Stress, 40 components, 211 normal, 211 shear, 211 tensor, 210, 212 vector, 9, 145 velocity gradient relations, 216 Strouhal, Vincenz, 246 Strouhal number, 246, 248–249, 257 Submerged body, 61 Submerged objects drag, 508 Subsonic flow, 430 Substantial derivative, 93 streamline coordinates, 215 Substantive derivatives in Cartesian, cylindrical and spherical coordinates, 96 Suction specific speed, 627 Supercavitation, 364 Superposition, 381–384 simple flows exercises dealing with, 417–419 Supersaturation, 455 Supersonic flow, 430 around convex corner, 462 large turning angles, 464 Supersonic nozzle, 434, 438 Surcharge, 549 Surface profiles classification, 515 Surface tension, 19–21 exercises dealing with, 35 Surge, 506 Swamee formula, 309 Swamee-Jain equation, 549 comparisons, 550 Swept-back airfoils, 370 System, 23, 128–129 to control volume, 133

exercises dealing with, 182 transformation, 132–136, 135 demand curve, 323, 629 pump characteristic curve, 629 example, 131 extensive property, 131 fixed control volume, 133

T Tangential force, 9 Target meter, 673–674 Taylor series expansion, 703 Temperature exercises dealing with, 32 Temperature scales, 11–13 Theoretical head curve turbopumps, 607 Thermal anemometer, 666 Thermal conductivity, 223 Thermal diffusivity, 225 Thermodynamic properties, 24 Thermodynamic quantities, 27 Thermodynamic relations, 427 Thoma cavitation number, 618 Thomas algorithm, 709 Three-dimensional flow, 100, 112 Three-dimensional imaging, 680–681 Three-dimensional stress components, 211 Time average, 293 mechanism, 659 streamlines, 348 velocity profile, 298 Time derivative control volume, 135 Torque, 291 Total head, 109, 148 Total pressure, 109 Tracers, 676–681 Trailing vortex, 370–371 Transition, 477 boundary layer, 386 region, 387 momentum equation, 502–503 zone, 307 equation, 307 Translating hydraulic jump, 506 Transportive property FD, 719 Trapezoidal channel example, 494, 525 Trapezoidal cross section, 486 Trapezoidal method, 708 Trapezoidal section, 481 Traveling cavitation, 363 Triangle areas, 743 Triangular channel example, 490 Troposphere, 44


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Index 793 Truncation error, 713 T-section energy equation, 150 Tufts, 678 Turbines, 634–649 exercises dealing with, 655–657 head, 150 meter, 672 rotor flow example, 169–170 selection and operation, 647–649 Turbomachinery, 601–650 cavitation, 615–619 dimensional analysis and similitude, 619–628 elementary theory exercises dealing with, 650–651 parameters, 620 similarity rules, 623–625 Turbopumps, 602–619 exercises dealing with, 652–654 in piping systems, 628–633 Turbulence intensity, 659 Turbulent boundary layers, 386–387, 393–396 empirical form, 397–402 exercises dealing with, 421–422 before separation, 353 velocity profile, 398 Turbulent flow, 103 differential equations, 229 empirical relations smooth pipe, 300 exercises dealing with, 330, 335–336 in pipe, 292–324, 295 exercises dealing with, 337–339 transition, 274 Turbulent profile, 387 Turbulent regime equation, 307 Turbulent region, 299 Turbulent shear stress, 295 Turbulent velocity profile, 302 Turbulent zone, 397 Two-dimensional flow, 100 Two-dimensional stress components, 211 Two-dimensional velocity profile, 499

U Uncertainty analysis categories, 684 flow measurements, 684–690 Uniform depth, 483 Uniform flow, 101, 477, 480 depth associated, 483 equation, 482–484 exercises dealing with, 531–532 turbulent in open channels, 325 in x-direction, 379

Unit normal vector, 132 Units, 4–8 exercises dealing with, 31 vector volume, 134 Unsteady flow, 477 exercises dealing with, 599 in pipelines, 576–584 Unswept airfoil, 370 U-tube manometer, 48

V Vacuum, 12 Valve closure rapid, 576–584 Vanes, 166 velocity, 604 Vapor flow exercises dealing with, 470 through nozzle, 454–456 Vapor pressure, 22, 111 exercises dealing with, 35 Varied flow function, 758–759 force, 9 stress, 9, 145 unit volume, 134 velocity diagram Pelton bucket, 642 Vector relationships, 736 Velocimetry, 677–678 Velocity area method, 666 coefficient, 643 components in turbulent pipe flow, 293 curves vs. pressure drop, 240 defect, 347, 397 distribution example, 155 of fluid particle, 92 gradients concept, 16 head, 148 laminar flow, 103 liquids and gases, 665 measurement, 661–666 measuring devices, 666 potential, 372 exercises dealing with, 416–417 profile, 298–305, 388 turbulent boundary layer, 398 in turbulent pipe flow, 274 propeller, 171 time intermittent flow, 105 turbulent flow, 103 vane, 604 vector diagram Pelton bucket, 642 Vena contractas, 317

Venturi meter, 667, 671 example, 154 flow coefficient K versus Reynolds number, 670 Venturi nozzle, 111 Vertical stability, 64 Vibration, 363 Vibratory cavitation, 363 Viscometer, 17 Viscosity, 15–18, 147, 205 exercises dealing with, 33–34 as function of temperature, 804 of oil, 291 second coefficient, 216 Viscous drag, 253 Viscous effects, 88, 111 boundary layer, 387 vorticity changes, 220 Viscous flow, 101 Viscous length, 299 Viscous shear, 296 Viscous wall layer, 299, 397 turbulence initiated, 299 V-notch weir, 497 Volumes properties, 743 unit vector, 134 Volute, 602, 635 Von Karman, Theodor, 359 Von Karman integral equation, 387 exercises dealing with, 420 Vortex cavitation, 363 Vortex flow, 383 Vortex line, 220 Vortex shedding, 359–362 exercises dealing with, 413–414 meter, 674–675 Vortex strength, 378, 381 Vortex tube, 220 Vorticity, 94–99 in Cartesian, cylindrical and spherical coordinates, 96 changes viscous effects, 220 defined, 95 equation, 373 equations, 219–222 exercises dealing with, 416–417

W Wake, 347 Wall region, 299, 301 Water hammer, 19, 576, 579 Water surface profiles, 514–517 numerical analysis, 520 Wave drag, 253 Weber, Moritz, 246 Weber number, 246, 248–249, 261 Wedge, 457 starting vortex, 221


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Index

Weir, 495 rectangle sharp-crested, 497 sharp-crested, 496 Weisbach, Julius, 278 Wetted perimeter, 313, 480 Wicket gates, 635 Wide channel, 285

Wind-tunnel airspeed prototype velocity, 256–257 Wingspan, 369 Work rate fluid element, 223 Work-rate term, 144–146

Y Yaw meter, 663

Z Zero-drag result, 382 Zero-lift prediction, 382 Zone of silence, 430


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Page 3

Properties of Water (English) Temperature (°F) 32 40 50 60 70 80 90 100 120 140 160 180 200 212

Density (slugs/ft3)

Specific weight (lb/ft3)

Viscosity (lb-sec/ft2)

Kinematic viscosity (ft2/sec)

1.94 1.94 1.94 1.94 1.94 1.93 1.93 1.93 1.92 1.91 1.90 1.88 1.87 1.86

62.4 62.4 62.4 62.4 62.4 62.1 62.1 62.1 61.8 61.5 61.1 60.5 60.2 59.8

3.75 ⫻ 10⫺5 3.23 2.74 2.36 2.05 1.80 1.60 1.42 1.17 0.98 0.84 0.73 0.64 0.59 ⫻ 10⫺5

1.93 ⫻ 10⫺5 1.66 1.41 1.22 1.06 0.93 0.826 0.739 0.609 0.514 0.442 0.385 0.341 0.319 ⫻ 10⫺5

Bulk modulus (lb/in2) 293,000 294,000 305,000 311,000 320,000 322,000 323,000 327,000 333,000 330,000 326,000 318,000 308,000 300,000

Surface tension (lb/ft) 0.518 ⫻ 10⫺2 0.514 0.509 0.504 0.500 0.492 0.486 0.480 0.465 0.454 0.441 0.426 0.412 0.404 ⫻ 10⫺2

Vapor pressure (lb/in2) 0.089 0.12 0.178 0.256 0.340 0.507 0.698 0.949 1.69 2.89 4.74 7.51 11.53 14.7

Properties of Air at Atmospheric Pressure (English) Temperature (°F) ⫺20 0 20 40 60 68 80 100 120 160 200 300 400 1000

Density (slugs/ft3) 0.00280 0.00268 0.00257 0.00247 0.00237 0.00233 0.00228 0.00220 0.00213 0.00199 0.00187 0.00162 0.00144 0.000844

Viscosity (lb-sec/ft2)

Kinematic viscosity (ft2/sec)

3.34 ⫻ 10⫺7 3.38 3.50 3.62 3.74 3.81 3.85 3.96 4.07 4.23 4.50 4.98 5.26 7.87 ⫻ 10⫺7

11.9 ⫻ 10⫺5 12.6 13.6 14.6 15.8 16.0 16.9 18.0 18.9 21.3 24.1 30.7 36.7 93.2 ⫻ 10⫺5

Speed of sound (ft/sec) 1028 1051 1074 1096 1117 1125 1138 1159 1180 1220 1258 1348 1431 1839


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